ML Aggarwal Solution Class 10 Chapter 18 Trigonometric Identities Test

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16

Test

Question 1

(i) If θ is an acute angle and cosec θ = √5 find the value of cot θ – cos θ.

(ii) If θ is an acute angle and $\tan \theta=\frac{8}{15}$ find the value of sec θ + cosec θ.

Sol :

(i) θ is an acute angle.

cosec θ = √5

$\therefore \sin \theta=\frac{1}{\sqrt{5}}$

$\cos \theta=\sqrt{1-\sin ^{2} \theta}$

$=\sqrt{1-\left(\frac{1}{\sqrt{5}}\right)^{2}}=\sqrt{1-\frac{1}{5}}$

$=\sqrt{\frac{5-1}{5}}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}$

Now, cotθ-cosθ$=\frac{\cos \theta}{\sin \theta}-\cos \theta$

$=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}-\frac{2}{\sqrt{5}}=\frac{2}{1}-\frac{2}{\sqrt{5}}$

$=2\left(1-\frac{1}{\sqrt{5}}\right)=\frac{2(\sqrt{5}-1)}{\sqrt{5}}$

(ii) $\tan \theta=\frac{8}{15}$

In the figure,

$\tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{8}{15}$

$\therefore \mathrm{BC}=8, \mathrm{AB}=15$

$\therefore \mathrm{AC}=\sqrt{\mathrm{AB}^{2}+\mathrm{BC}^{2}}=\sqrt{5^{2}+8^{2}}$

$=\sqrt{225+64}=\sqrt{289}=17$

$\therefore \sec \theta=\frac{A C}{A B}=\frac{17}{15}$

$\operatorname{cosec} \theta=\frac{A C}{B C}=\frac{17}{8}$

$\therefore \sec \theta+\operatorname{cosec} \theta=\frac{17}{15}+\frac{17}{8}$

$=\frac{136+255}{120}=\frac{391}{120}=3 \frac{31}{120}$

Question 2

Evaluate the following:

(i) $2 \times\left(\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 25^{\circ}+\sin ^{2} 65^{\circ}}\right)$– tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

(ii) $\frac{\sec 29^{\circ}}{\operatorname{cosec} 61^{\circ}}$+ 2 cot 8° cot 17° cot 45° cot 73°0 cot 82° – 3(sin2 38° + sin2 52°)

(iii) $\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}$+ $sin^2$ 63° + cos 63° sin 27°

Sol :

(i) $2 \times\left(\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 25^{\circ}+\sin ^{2} 65^{\circ}}\right)$– tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

$=2\left[\frac{\cos ^{2}\left(90^{\circ}-70^{\circ}\right)+\cos ^{2} 70^{\circ}}{\sin ^{2} 25^{\circ}+\sin ^{2}\left(90^{\circ}-25^{\circ}\right)}\right]-\tan 45^{\circ}$$+\tan 13^{\circ} \tan 77^{\circ} \tan 23^{\circ} \tan 67^{\circ} \tan 30^{\circ}$

$=2\left[\frac{\sin ^{2} 70^{\circ}+\cos ^{2} 70}{\sin ^{2} 25^{\circ}+\cos ^{2} 25^{\circ}}\right]-1+\tan 13^{\circ} \tan$ $-1+\tan 13^{\circ} \tan$

$\left(90^{\circ}-13^{\circ}\right) \tan 23^{\circ} \tan \left(90^{\circ}-23^{\circ}\right) \times \frac{1}{\sqrt{3}}$

$=2\left(\frac{1}{1}\right)-1+\tan 13 \cot 13^{\circ} \tan 23^{\circ} \cot 23^{\circ}\times \frac{1}{\sqrt{3}}$

$=2-1+1 \times 1 \times \frac{1}{\sqrt{3}}$

$=2-1+\frac{1}{\sqrt{3}}^{\cdot}$

$=1+\frac{1}{\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{3}}$

$=\frac{(\sqrt{3}+1) \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$=\frac{3+\sqrt{3}}{3}$

(ii) $\frac{\sec 29^{\circ}}{\operatorname{cosec} 61^{\circ}}+2 \cot 8^{\circ} \cot 17^{\circ} \cot 45^{\circ} \cot$ 

$73^{\circ} \cot 82^{\circ}-3\left(\sin ^{2} 38^{\circ}+\sin ^{2} 52^{\circ}\right)$

$=\frac{\sec 29^{\circ}}{\operatorname{cosec}\left(90^{\circ}-29^{\circ}\right)}+2 \cot 8^{\circ} \times \cot \left(90^{\circ}-\right.\left.8^{\circ}\right) \times \cot 17^{\circ} \times \cot \left(90^{\circ}-17^{\circ}\right) \cot 45^{\circ}-3\left[\sin ^{2} 38^{\circ}+\sin ^{2}\left(90^{\circ}-38^{\circ}\right)\right]$

$=\frac{\sec 29^{\circ}}{\sec 29^{\circ}}+2 \cot 8^{\circ} \tan 8^{\circ} \times \cot 17^{\circ} \tan 17^{\circ} \times 1-3\left(\sin ^{2} 38^{\circ}+\cos ^{2} 38^{\circ}\right)$

$=1+2 \times 1 \times 1 \times 1-3 \times 1=1+2-3=0$

(iii) $\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}$

$=\frac{\sin ^{2} 22^{\circ}+\sin ^{2}\left(90^{\circ}-22^{\circ}\right)}{\cos ^{2} 22^{\circ}+\cos ^{2}\left(90^{\circ}-22^{\circ}\right)}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin \left(90^{\circ}-63^{\circ}\right)$

$=\frac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\cos ^{2} 22^{\circ}+\sin ^{2} 22^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ}\times \cos 63^{\circ}$

$=\frac{1}{1}+\left(\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ}\right)$

=1+1=2

Question 3

If $\frac{4}{3}\left(\sec ^259^{\circ}-\cot ^231^{\circ}\right)-\frac{2}{2} \sin 90^{\circ}+3 \tan^256^{\circ} \tan ^234^{\circ}=\frac{x}{2}$then find the value of x.

Sol :

Given

$\frac{4}{3}\left(\sec ^259^{\circ}-\cot ^231^{\circ}\right)-\frac{2}{2} \sin 90^{\circ}+3 \tan^256^{\circ} \tan ^234^{\circ}=\frac{x}{2}$

$\Rightarrow \frac{4}{3}\left[\sec ^{2} 59^{\circ}-\cot ^{2}\left(90^{\circ}-59^{\circ}\right)\right]-\frac{2}{3} \sin 90^{\circ}+3 \tan ^{2} 56^{\circ} \tan ^{2}\left(90^{\circ}-56^{\circ}\right)=\frac{x}{3}$

$=\frac{4}{3}\left[\sec ^{2} 59^{\circ}-\tan ^{2} 59^{\circ}\right]-\frac{2}{3} \times 1+3\tan ^{2} 56^{\circ} \cot ^{2} 56^{\circ}=\frac{x}{3}$

$=\frac{4}{3} \times 1-\frac{2}{3}+3 \times 1=\frac{x}{3}$

$\Rightarrow \frac{4}{3}-\frac{2}{3}+3=\frac{x}{3}$

$\Rightarrow \frac{4-2+9}{3}=\frac{x}{3}$

$\Rightarrow \frac{11}{3}=\frac{x}{3}$

$ \Rightarrow x=\frac{11 \times 3}{3}=11$

∴x=11

Prove the following (4 to 11) identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

Question 4

(i) $\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}=2 \sec A$

(ii) $\frac{\cos A}{\operatorname{cosec} A+1}+\frac{\cos A}{\operatorname{cosec} A-1}=2 \tan A$

Sol :

(i) $\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}=2 \sec A$

L.H.S$=\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}$

$=\cos \mathrm{A}\left[\frac{1}{1-\sin \mathrm{A}}+\frac{1}{1+\sin \mathrm{A}}\right]$

$=\cos \mathrm{A}\left[\frac{1+\sin \mathrm{A}+1-\sin \mathrm{A}}{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}\right]$

$=\cos \mathrm{A}\left[\frac{2}{1-\sin ^{2} \mathrm{~A}}\right]=\frac{2 \cos \mathrm{A}}{\cos ^{2} \mathrm{~A}}$

$=\frac{2}{\cos A}=2 \sec A$=R.H.S

(ii) $\frac{\cos A}{\operatorname{cosec} A+1}+\frac{\cos A}{\operatorname{cosec} A-1}=2 \tan A$

L.H.S$=\frac{\cos A}{\operatorname{cosec} A+1}+\frac{\cos A}{\operatorname{cosec} A-1}$

$=\cos A\left[\frac{1}{\operatorname{cosec} A+1}+\frac{1}{\operatorname{cosec} A-1}\right]$

$=\cos A\left[\frac{\operatorname{cosec} A-1+\operatorname{cosec} A+1}{(\operatorname{cosec} A+1)(\operatorname{cosec} A-1)}\right]$

$=\frac{\cos A[2 \operatorname{cosec} A]}{\operatorname{cosec}^{2} A-1}=\frac{2 \cos A}{\sin A\left(\cot ^{2} A\right)}$

$=\frac{2 \cot A}{\cot ^{2} A}=\frac{2}{\cot A}$

=2 tan A= R.H.S

Question 5

(i) $\frac{(\cos \theta-\sin \theta)(1+\tan \theta)}{2 \cos ^{2} \theta-1}=\sec \theta$

(ii) (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1.

Sol :

(i) $\frac{(\cos \theta-\sin \theta)(1+\tan \theta)}{2 \cos ^{2} \theta-1}=\sec \theta$

L.H.S $=\frac{(\cos \theta-\sin \theta)(1+\tan \theta)}{2 \cos ^{2} \theta-1}$

$=\frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$

$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$

$=\frac{\cos ^{2} \theta-\left(1-\cos ^{2} \theta\right)}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$

$=\frac{\cos ^{2} \theta-1+\cos ^{2} \theta}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$

$=\frac{2 \cos ^{2} \theta-1}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$

$=\frac{1}{\cos \theta}=\sec \theta$=R.H.S

(ii) $(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)=1$

$(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)=1$

L.H.S. $=(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$

$=\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta\right)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$

$=\left(\frac{1-\sin ^{2} \theta}{\sin \theta}\right)\left(\frac{1-\cos ^{2} \theta}{\cos \theta}\right)\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}\right)$

$=\frac{\left(\cos ^{2} \theta \times \sin ^{2} \theta\right)}{\sin \theta \cos \theta} \times \frac{1}{\sin \theta \cos \theta}$

$=\frac{\sin ^{2} \cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}=1$=R.H.S

Question 6

(ii) $\sin ^{2} \theta+\cos ^{4} \theta=\cos ^{2} \theta+\sin ^{4} \theta$

(ii) $\frac{\cot \theta}{\operatorname{cosec} \theta+1}+\frac{\operatorname{cosec} \theta+1}{\cot \theta}=2 \sec \theta$

Sol :

L.H.S$=\sin ^{2} \theta+\cos ^{4} \theta$

$=\left(1-\cos ^{2} \theta+\cos ^{4} \theta\right.$

$=1-\cos ^{2} \theta+\cos ^{4} \theta$

$=1-\cos ^{2} \theta\left(1-\cos ^{2} \theta\right)$

R.H.S$=\cos ^{2} \theta+\sin ^{4} \theta$

$=1-\sin ^{2} \theta+\sin ^{4} \theta$

$=1-\sin ^{2} \theta\left(1-\sin ^{2} \theta\right)$

$=1-\sin ^{2} \theta \cos ^{2} \theta$

∴L.H.S=R.H.S

(ii) $\frac{\cot \theta}{\operatorname{cosec} \theta+1}+\frac{\operatorname{cosec} \theta+1}{\cot \theta}=2 \sec \theta$

L.H.S$=\frac{\cot \theta}{\operatorname{cosec} \theta+1}+\frac{\operatorname{cosec} \theta+1}{\cot \theta}$

$=\frac{\frac{\cos \theta}{\sin \theta}}{\frac{1}{\sin \theta}+1}+\frac{\frac{1}{\sin \theta}+1}{\frac{\cos \theta}{\sin \theta}}$

$=\frac{\frac{\cos \theta}{\sin \theta}}{\frac{1+\sin \theta}{\sin \theta}}+\frac{\frac{1+\sin \theta}{\sin \theta}}{\frac{\cos \theta}{\sin \theta}}$

$=\frac{\cos \theta}{\sin \theta} \times \frac{\sin \theta}{1+\sin \theta}+\frac{1+\sin \theta}{\sin \theta} \times \frac{\sin \theta}{\cos \theta}$

$=\frac{\cos \theta}{1+\sin \theta}+\frac{1+\sin \theta}{\cos \theta}$

$=\frac{\cos ^{2} \theta+1+\sin ^{2} \theta+2 \sin \theta}{(1+\sin \theta) \cos \theta}$

$=\frac{1+1+2 \sin \theta}{(1+\sin \theta) \cos \theta}=\frac{2+2 \sin \theta}{(1+\sin \theta) \cos \theta}$

$=\frac{2(1+\sin \theta)}{\cos \theta(1+\sin \theta)}=\frac{2}{\cos \theta}$

=2 sec θ=R.H.S

Question 7

(i) $\sec ^{4} A\left(1-\sin ^{4} A\right)-2 \tan ^{2} A=1$

(ii) $\frac{1}{\sin A+\cos A+1}+\frac{1}{\sin A+\cos A-1}=\sec A+\operatorname{cosec} A$

Sol :

(i) $\sec ^{4} \mathrm{~A}\left(1-\sin ^{4} \mathrm{~A}\right)-2 \tan ^{2} \mathrm{~A}=1$

L.H.S$=\sec ^{4} A\left(1-\sin ^{4} A\right)-2 \tan ^{2} A$

$=\frac{1}{\cos ^{4} A}\left(1+\sin ^{2} A\right)\left(1-\sin ^{2} A\right)-2 \tan ^{2} A$

$\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$

$=\frac{\left(1+\sin ^{2} A\right) \cos ^{2} A}{\cos ^{4} A}-2 \frac{\sin ^{2} A}{\cos ^{2} A}$

$=\frac{1+\sin ^{2} A}{\cos ^{2} A}-\frac{2 \sin ^{2} A}{\cos ^{2} A}$

$=\frac{1+\sin ^{2} A-2 \sin ^{2} A}{\cos ^{2} A}$

$=\frac{1-\sin ^{2} A}{\cos ^{2} A}$

$=\frac{\cos ^{2} A}{\cos ^{2} A}=1$

$\left(\because 1-\sin ^{2} A=\cos ^{2} A\right)$

=R.H.S

(ii) $\frac{1}{\sin A+\cos A+1}+\frac{1}{\sin A+\cos A-1}$

=sec A+cosec A

L.H.S$=\frac{1}{\sin A+\cos A+1}+\frac{1}{\sin A+\cos A-1}$

$=\frac{\sin A+\cos A-1+\sin A+\cos A+1}{(\sin A+\cos A+1)(\sin A+\cos A-1)}$

$=\frac{2(\sin A+\cos A)}{(\sin A+\cos A)^{2}-(1)^{2}}$

$=\frac{2(\sin A+\cos A)}{\sin ^{2} A+\cos ^{2} A+2 \sin A \cos A-1}$

$=\frac{2(\sin A+\cos A)}{1+2 \sin A \cos A-1}$

$=\frac{2(\sin A+\cos A)}{2 \sin A \cos A}$

$=\frac{\sin A+\cos A}{\sin A \cos A}$

$=\frac{\sin A}{\sin A \cos A}+\frac{\cos A}{\sin A \cos A}$

$=\frac{1}{\cos A}+\frac{1}{\sin A}$

=sec A+cosec A=R.H.S

Question 8

(i) $\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta \cos \theta}+\sin \theta \cos \theta=1$

(ii) $(\sec A-\tan A)^{2}(1+\sin A)=1-\sin A$

Sol :

(i) $\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta \cos \theta}+\sin \theta \cos \theta=1$

L.H.S$=\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta \cos \theta}+\sin \theta \cos \theta$

$=\frac{(\sin \theta+\cos \theta)\left(\sin ^{2} \theta-\sin \theta \cos \theta+\cos ^{2} \theta\right)}{(\sin \theta+\cos \theta)}$$+\sin \theta \cos \theta$

$=\sin ^{2} \theta+\cos ^{2} \theta-\sin \theta \cos \theta+\sin \theta \cos \theta$

=1=R.H.S

(ii) $(\sec A-\tan A)^{2}(1+\sin A)^{\doteq}=1-\sin A$

L.H.S$=\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)^{2}(1+\sin A)$

$=\left(\frac{1-\sin \mathrm{A}}{\cos \mathrm{A}}\right)^{2}(1+\sin \mathrm{A})$

$=\frac{(1-\sin A)^{2}}{\cos ^{2} A}(1+\sin A)$

$=\frac{(1-\sin \mathrm{A})^{2}(1+\sin \mathrm{A})}{1-\sin ^{2} \mathrm{~A}}$

$=\frac{(1-\sin \mathrm{A})^{2}(1+\sin \mathrm{A})}{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}$

=1-sin A=R.H.S

Question 9

(i) $\frac{\cos A}{1-\tan A}-\frac{\sin ^{2} A}{\cos A-\sin A}=\sin A+\cos A$

(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A

(iii) $\frac{\tan ^{2} \theta}{\tan ^{2} \theta-1}-\frac{\operatorname{cosec}^{2} \theta}{\sec ^{2} \theta-\operatorname{cosec}^{2} \theta}=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}$

Sol :

(i) $\frac{\cos A}{1-\tan A}-\frac{\sin ^{2} A}{\cos A-\sin A}=\sin A+\cos A$

L.H.S$=\frac{\cos A}{1-\tan A}-\frac{\sin ^{2} A}{\cos A-\sin A}$

$=\frac{\cos A}{1-\frac{\sin A}{\cos A}}-\frac{\sin ^{2} A}{\cos A-\sin A}$

$=\frac{\cos A \times \cos A}{\cos A-\sin A}-\frac{\sin ^{2} A}{\cos A-\sin A}$

$=\frac{\cos ^{2} A}{\cos A-\sin A}-\frac{\sin ^{2} A}{\cos A-\sin A}$

$=\frac{\cos ^{2} A-\sin ^{2} A}{\cos A-\sin A}$

$=\frac{(\cos A+\sin A)(\cos A-\sin A)}{(\cos A-\sin A)}$

=cos A+sin A

=sin A+cos A=R.H.S

(ii) $(\sec A-\operatorname{cosec} A)(1+\tan A+\cot A)=\tan A\sec A-\cot A \operatorname{cosec} A$

L.H.S=(secA-cosecA)(1+tanA+cotA)

$=\left(\frac{1}{\cos A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)$

$=\frac{\sin A-\cos A}{\sin A \cos A}\times \frac{\sin A \cos A+\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}$

$=\frac{\sin A-\cos A}{\sin A \cos A} \times \frac{\sin A \cos A+1}{\sin A \cos A}$

$=\frac{(\sin A-\cos A)(\sin A \cos A+1)}{\sin ^{2} A \cos ^{2} A}$

$=\frac{\sin A}{\cos A} \cdot \frac{1}{\cos A}-\frac{\cos A}{\sin A} \cdot \frac{1}{\sin A}$

$=\frac{\sin A}{\cos ^{2} A}-\frac{\cos A}{\sin ^{2} A}$

$=\frac{\sin ^{3} A-\cos ^{3} A}{\sin ^{2} A \cos ^{2} A}$

$=\frac{(\sin A-\cos A)\left(\sin ^{2} A+\cos ^{2} A+\sin A \cos A\right)}{\sin ^{2} A \cos ^{2} A}$

$=\frac{(\sin \mathrm{A}-\cos A)(1+\sin \mathrm{A} \cos \mathrm{A})}{\sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~A}}$

$=\frac{(\sin \mathrm{A}-\cos \mathrm{A})(\sin \mathrm{A} \cos \mathrm{A}+1)}{\sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~A}}$

∴L.H.S=R.H.S

(iii) $\frac{\tan ^{2} \theta}{\tan ^{2} \theta-1}+\frac{\operatorname{cosec}^{2} \theta}{\sec ^{2} \theta-\operatorname{cosec}^{2} \theta}=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}$

L.H.S$=\frac{\tan ^{2} \theta}{\tan ^{2} \theta-1}+\frac{\operatorname{cosec}^{2} \theta}{\sec ^{2} \theta-\operatorname{cosec}^{2} \theta}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}-1}+\frac{\frac{1}{\sin ^{2} \theta}}{\frac{1}{\cos ^{2} \theta}-\frac{1}{\sin ^{2} \theta}}$

$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}+\frac{\frac{1}{\sin ^{2} \theta}}{\frac{\sin ^{2} \theta-\cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}}$

$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}+\frac{\sin ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta\left(\sin ^{2} \theta-\cos ^{2} \theta\right)}$

$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}+\frac{\cos ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}$

$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}$

=R.H.S

Question 10

$\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{\sin ^{2} A-\cos ^{2} A}=\frac{2}{1-2 \cos ^{2} A}=\frac{2 \sec ^{2} A}{\tan ^{2} A-1}$

Sol :

$\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{\sin ^{2} A-\cos ^{2} A}=\frac{2}{1-2 \cos ^{2} A}=\frac{2 \sec ^{2} A}{\tan ^{2} A-1}$

L.H.S$=\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}$

$=\frac{(\sin \mathrm{A}+\cos \mathrm{A})^{2}+(\sin \mathrm{A}-\cos \mathrm{A})^{2}}{\sin ^{2} \mathrm{~A}-\cos ^{2} \mathrm{~A}}$

$=\frac{2\left(\sin ^{2} A+\cos ^{2} A\right)}{\sin ^{2} A-\cos ^{2} A}$

$=\frac{2 \times 1}{\sin ^{2} A-\cos ^{2} A}=\frac{2}{\sin ^{2} A-\cos ^{2} A}$

=R.H.S

Now, we have

$=\frac{2}{\sin ^{2} A-\cos ^{2} A}$

$=\frac{2}{\sin ^{2} A-\left(1-\sin ^{2} A\right)}$

$=\frac{2}{\sin ^{2} A-1+\sin ^{2} A}$

$=\frac{2}{\sin ^{2} A-\cos ^{2} A}$

$=\frac{\frac{2}{\cos ^{2} A}}{\frac{\sin ^{2} A}{\cos ^{2} A}-\frac{\cos ^{2} A}{\cos ^{2} A}}$

$=\frac{2 \sec ^{2} A}{\tan ^{2} A-1}$

Question 11

$2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1=\theta$

Sol :

$2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1=\theta$

L.H.S=$=2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1$

$=2\left[\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}\right]-3\left[\sin ^{4} \theta+\cos ^{4} \theta\right]+1$

$=2\left[\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\right.\left.\left(\sin ^{4} \theta+\cos ^{4} \theta-\sin ^{2} \theta \cos ^{2} \theta\right)\right]-3\left[\sin ^{4} \theta+\cos ^{4} \theta\right]+1$

$=2\left(\sin ^{4} \theta+\cos ^{4} \theta-\sin ^{2} \theta \cos ^{2} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1$

$=-2 \sin ^{2} \theta \cos ^{2} \theta-\sin ^{4} \theta-\cos ^{4} \theta+1$

$=1-\left[\sin ^{4} \theta+\cos ^{4} \theta+2 \sin ^{2} \theta \cos ^{2} \theta\right]$

$=1-\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}=1-1=0$

=R.H.S

Question 12

If cot θ + cos θ = m, cot θ – cos θ = n, then prove that $\left(m^{2}-n^{2}\right)^{2}$= 16 run.

Sol :

cot θ + cos θ = m…..(i)

cot θ – cos θ = n……(ii)

Adding (i)&(ii) we get

$2 \cot \theta=m+n $

$\Rightarrow \cot \theta=\frac{m+n}{2}$

$\therefore \tan \theta=\frac{2}{m+n}$...(i)

Subtracting (ii) from (i)

$2 \cos \theta=m-n \Rightarrow \cos \theta=\frac{m-n}{2}$

$\therefore \quad \sec \theta=\frac{2}{m-n}$...(ii)

Squaring and subtracting

$\sec ^{2} \theta-\tan ^{2} \theta=\left(\frac{2}{m-n}\right)^{2}-\left(\frac{2}{m+n}\right)^{2}$

$1=\frac{4}{(m-n)^{2}}-\frac{4}{(m+n)^{2}}$

$\Rightarrow 4\left[\frac{1}{(m-n)^{2}}-\frac{1}{(m+n)^{2}}\right]=1$

$\Rightarrow 4\left[\frac{(m+n)^{2}-(m-n)^{2}}{(m+n)^{2}(m-n)^{2}}\right]=1$

$\Rightarrow \frac{4(4 m n)}{\left(m^{2}-n^{2}\right)^{2}}=1$

$\Rightarrow \frac{16 m n}{\left(m^{2}-n^{2}\right)^{2}}=1$

$\Rightarrow \quad\left(m^{2}-n^{2}\right)^{2}=16 \mathrm{mn}$

Question 13

If sec θ + tan θ = p, prove that sin θ = $\frac{p^{2}-1}{p^{2}+1}$

Sol :

sec θ + tan θ = p,

prove that sin θ $=\frac{p^{2}-1}{p^{2}+1}$

$\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}=p$

$\frac{1+\sin \theta}{\cos \theta}=p$

Squaring ,

$\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}=p^{2}$

$ \Rightarrow \frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta}=p^{2}$

$\Rightarrow \frac{(1+\sin \theta)^{2}}{(1-\sin \theta)(1+\sin \theta)}=p^{2}$

$\Rightarrow \frac{1+\sin \theta}{1-\sin \theta}=\frac{p^{2}}{1}$

Applying componendo and dividendo

$\frac{1+\sin \theta+1-\sin \theta}{1+\sin \theta-1+\sin \theta}=\frac{p^{2}+1}{p^{2}-1}$

$\frac{1+\sin \theta+1-\sin \theta}{1+\sin \theta-1+\sin \theta}$

$=\frac{p^{2}+1}{p^{2}-1}$

$\therefore \sin \theta=\frac{p^{2}-1}{p^{2}+1}$

Question 14

If tan A = n tan B and sin A = m sin B, prove that $\cos ^{2} A=\frac{m^{2}-1}{n^{2}-1}$

Sol :

$m=\frac{\sin A}{\sin B}$

$n=\frac{\tan A}{\tan B}$

$\frac{1}{\sin \mathrm{B}}=\frac{m}{\sin \mathrm{A}}$

$ \Rightarrow \operatorname{cosec} \mathrm{B}=\frac{m}{\sin \mathrm{A}}$

$\frac{1}{\tan \mathrm{B}}=\frac{n}{\tan \mathrm{A}} $

$\Rightarrow \cot \mathrm{B}=\frac{n}{\tan \mathrm{A}}$

Now, $\operatorname{cosec}^{2} B-\cot ^{2} B=1$

$\Rightarrow \frac{m^{2}}{\sin ^{2} A}-\frac{n^{2}}{\tan ^{2} A}=1$

$\Rightarrow \frac{m^{2}}{\sin ^{2} A}-\frac{n^{2} \cos ^{2} A}{\sin ^{2} A}=1$

$\Rightarrow m^{2}-n^{2} \cos ^{2} A=\sin ^{2} A$

$\Rightarrow m^{2}-n^{2} \cos ^{2} A=1-\cos ^{2} A$

$\Rightarrow m^{2}-1=\cos ^{2} A+n^{2} \cos ^{2} A$

$\Rightarrow m^{2}-1=\left(n^{2}-1\right) \cos ^{2} A$

$\Rightarrow \cos ^{2} A=\frac{m^{2}-1}{n^{2}-1}$

Question 15

If sec A  $=x+\frac{1}{4 x}$, , then prove that sec A + tan A = 2x or $\frac{1}{2 x}$

Sol :

$\sec A=x+\frac{1}{4 x}$

To prove that sec A + tan A = 2x or $\frac{1}{2 x}$

$\tan A=\sqrt[\pm]{\sec ^{2} A-1}$

$=\sqrt[\pm]{\left(x+\frac{1}{4 x}\right)^{2}-1}$

$=\sqrt[\pm]{x^{2}+\frac{1}{16 x^{2}}+\frac{1}{2}-1}$

$=\sqrt[\pm]{x^{2}+\frac{1}{16 x^{2}}-\frac{1}{2}}$

$=\pm\left(x-\frac{1}{4 x}\right)$

$\therefore \sec A+\tan A=x+\frac{1}{4 x}+x-\frac{1}{4 x}=2 x$

or $x+\frac{1}{4 x}-x+\frac{1}{4 x}=\frac{1}{2 x}$

Hence proved

Question 16

When 0° < θ < 90°, solve the following equations:

(i) $2 \cos ^{2} \theta+\sin \theta-2=0$

(ii) $3 \cos \theta=2 \sin ^{2} \theta$

(iii) $\sec ^{2} \theta-2 \tan \theta=0$

(iv) $\tan ^{2} \theta=3(\sec \theta-1)$

Sol :

0° < θ < 90°

(i) $2 \cos ^{2} \theta+\sin \theta-2=0$

$2\left(1-\sin ^{2} \theta\right)+\sin \theta-2=0$

$2-2 \sin ^{2} \theta+\sin \theta-2=0$

$-2 \sin ^{2} \theta+\sin \theta=0$

$\Rightarrow \sin \theta(-2 \sin \theta+1)=0$

Either $\sin \theta=0,$ then $\theta=0$

or $-2 \sin \theta+1=0,$ then $\sin \theta=\frac{1}{2}$

$\therefore \theta=30^{\circ}$

Hence $\theta=0^{\circ}$ or $30^{\circ}$

(ii) $3 \cos \theta=2 \sin ^{2} \theta$

$\Rightarrow 3 \cos \theta=2\left(1-\cos ^{2} \theta\right)$

$\Rightarrow 3 \cos \theta=2-2 \cos ^{2} \theta$

$\Rightarrow 2 \cos ^{2} \theta+3 \cos \theta-2=0$

$\Rightarrow 2 \cos ^{2} \theta+4 \cos \theta-\cos \theta-2=0$

$\Rightarrow 2 \cos \theta(\cos \theta+2)-1(\cos \theta+2)=0$

$(\cos \theta+2)(2 \cos \theta-1)=0$

Either $\cos \theta+2=0 \Rightarrow \cos \theta=-2$ which is not possible being negative.

or $2 \cos \theta-1=0, \cos \theta=\frac{1}{2}=\cos 60^{\circ}$

$\therefore \theta=60^{\circ}$

(iii) $\sec ^{2} \theta-2 \tan \theta=0$

$1+\tan ^{2} \theta-2 \tan \theta=0$

$\Rightarrow \tan ^{2} \theta-2 \tan \theta+1=0$

$(\tan \theta-1)^{2}=0 \Rightarrow \tan \theta-1=0$

$\Rightarrow \tan \theta=1=\tan 45^{\circ}$

$\therefore \theta=45^{\circ}$

(iv) $\tan ^{2} \theta=3(\sec \theta-1)$

$\left(\sec ^{2} \theta-1\right)=3(\sec \theta-1)$

$\Rightarrow \sec ^{2} \theta-1=3 \sec \theta-3$

$\Rightarrow \sec ^{2} \theta-1-3 \sec \theta+3=0$

$\Rightarrow \sec ^{2} \theta-3 \sec \theta+2=0$

$\Rightarrow \sec ^{2} \theta-\sec \theta-2 \sec \theta+2=0$

$\Rightarrow \sec \theta(\sec \theta-1)-2(\sec \theta-1)=0$

$\Rightarrow(\sec \theta-1)(\sec \theta-2)=0$

Either $\sec \theta-1=0, \sec \theta=1=\sec 0^{\circ}$

$\therefore \theta=0^{\circ}$ not possible $\left(\because \theta>0^{\circ}\right)$

or $\sec \theta-2=0 \Rightarrow \sec \theta=2$

$\therefore \sec \theta=2=\sec 60^{\circ}$

$\therefore \theta=60^{\circ}$

ML Aggarwal Solution Class 10 Chapter 18 Trigonometric Identities MCQs

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

MCQs 

Choose the correct answer from the given four options (1 to 12) :

Question 1

$\cot ^{2} \theta-\frac{1}{\sin ^{2} \theta}$ is equal to

(a) 1

(b) -1

(c) $\sin ^{2} \theta$

(d) $\sec ^{2} \theta$

Sol :

$\cot ^{2} \theta-\frac{1}{\sin ^{2} \theta}$

$=\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-\frac{1}{\sin ^{2} \theta}$

$\frac{\cos ^{2} \theta-1}{\sin ^{2} \theta}=\frac{-\sin ^{2} \theta}{\sin ^{2} \theta}=-1$

Ans (b)

Question 2

$\left(\sec ^{2} \theta-1\right)\left(1-\operatorname{cosec}^{2} \theta\right)$ is equal to

(a) – 1

(b) 1

(c) 0

(d) 2

Sol :

$\left(\sec ^{2} \theta-1\right)\left(1-\operatorname{cosec}^{2} \theta\right)$

$=\left(\frac{1}{\cos ^{2} \theta}-1\right)\left(1-\frac{1}{\sin ^{2} \theta}\right)$

$=\frac{1-\cos ^{2} \theta}{\cos ^{2} \theta} \times \frac{\sin ^{2} \theta-1}{\sin ^{2} \theta}$

$=\frac{-\sin ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}=-1$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

Ans (a)

Question 3

$\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}$ is equal to

(a) $2 \sin ^{2} \theta$

(b) $2 \cos ^{2} \theta$

(c) $\sin ^{2} \theta$

(d) $\cos ^{2} \theta$

Sol :

$\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}$

$\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{1+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos ^{2} \theta}}$

$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta+\cos ^{2} \theta}$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

$=\frac{\sin ^{2} \theta}{1}=\sin ^{2} \theta$

Ans (c)

Question 4

$(\cos \theta+\sin \theta)^{2}+(\cos \theta-\sin \theta)^{2}$ is equal to

(a) – 2

(b) 0

(c) 1

(d) 2

Sol :

$(\cos \theta+\sin \theta)^{2}+(\cos \theta-\sin \theta)^{2}$

$=\cos ^{2} \theta+\sin ^{2} \theta+2 \sin \theta \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta-2 \sin \theta \cos \theta$

$=2\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$

$=2 \times 1=2(d)$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

Question 5

(sec A + tan A) (1 – sin A) is equal to

(a) sec A

(b) sin A

(c) cosec A

(d) cos A

Sol :

(sec A + tan A) (1 – sin A)

$=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)$

$=\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}} \times 1-\sin \mathrm{A}$

$=\frac{(1+\sin \mathrm{A})(1-\sin \mathrm{A})}{\cos \mathrm{A}}$

$=\frac{1-\sin ^{2} A}{\cos A}=\frac{\cos ^{2} A}{\cos A}=\cos A$

Ans (d)

Question 6

$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$ is equal to

(a) $\sec ^{2} A$

(b) -1

(c) $\cot ^{2} A$

(d) $\tan ^{2} A$

Sol :

$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$

$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}$

$=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}}$

$=\frac{1}{\cos ^{2} A} \times \frac{\sin ^{2} A}{1}$

$=\frac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A$

Ans (d)

Question 7

If sec θ – tan θ = k, then the value of sec θ + tan θ is

(a) $1-\frac{1}{k}$

(b) $1-k$

(c) $1+k$

(d) $\frac{1}{k}$

Sol :

sec θ – tan θ = k

$\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}=k$

$\frac{1-\sin \theta}{\cos \theta}=k$

Squaring both sides, we get

$\left(\frac{1-\sin \theta}{\cos \theta}\right)^{2}=(k)^{2} \Rightarrow \frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}=k^{2}$

$=\frac{(1-\sin \theta)^{2}}{1-\sin ^{2} \theta}=k^{2}$

$\Rightarrow \frac{(1-\sin \theta)^{2}}{(1+\sin \theta)(1-\sin \theta)}=k^{2}$

$=\frac{1-\sin \theta}{1+\sin \theta}=k^{2}$

$\Rightarrow \frac{1+\sin \theta}{1-\sin \theta}=\frac{1}{k^{2}}$

$ \Rightarrow \sqrt{\frac{(1+\sin \theta)}{1-\sin \theta}}=\frac{1}{k}$

Multiplying and dividing by $(1+\sin \theta)$

$=\sqrt{\left[\frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta}\right]}=\frac{1}{k}$

$\Rightarrow \sqrt{\left[\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}\right]}=\frac{1}{k}$

$\frac{1+\sin \theta}{\cos \theta}=\frac{1}{k}=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}=\frac{1}{k}$

$\Rightarrow \sec \theta+\tan \theta=\frac{1}{k}$

Ans (d)

Question 8

Which of the following is true for all values of θ (0° < θ < 90°):

(a) $\cos ^{2} \theta-\sin ^{2} \theta=1$

(b) $\operatorname{cosec}^{2} \theta-\sec ^{2} \theta=1$

(c) $\sec ^{2} \theta-\tan ^{2} \theta=1$

(c) $\cot ^{2} \theta-\tan ^{2} \theta=1$

Sol :

$\therefore \sec ^{2} \theta-\tan ^{2} \theta=1$ is true for all values of θ as it is an identity.

(0° < θ < 90°)

Ans (c)

Question 9

If θ is an acute angle of a right triangle, then the value of sin θ cos (90° – θ) + cos θ sin (90° – θ) is

(a) 0

(b) 2 sin θ cos θ

(c) 1

(c) $2 \sin ^{2} \theta$

Sol :

sin θ cos (90° – θ) + cos θ sin (90° – θ)

= sin θ sin θ + cos θ cos θ

{ ∵ sin(90° – θ) = cosθ, cos (90° – θ) = sin θ }

$=\sin ^{2} \theta+\cos ^{2} \theta=1$

Ans (c)

Question 10

The value of cos 65° sin 25° + sin 65° cos 25° is

(a) 0

(b) 1

(b) 2

(d) 4

Sol :

cos 65° sin 25° + sin 65° cos 25°

= cos (90° – 25°) sin 25° + sin (90° – 25°) cos 25°

= sin 25° . sin 25° + cos 25° . cos 25°

$=\sin ^{2} 25^{\circ}+\cos ^{2} 25^{\circ}$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

=1

Ans (b)

Question 11

The value of $3 \tan ^{2} 26^{\circ}-3 \operatorname{cosec}^{2} 64^{\circ}$ is

(a) 0

(b) 3

(c) – 3

(d) – 1

Sol :

$3 \tan ^{2} 26^{\circ}-3 \operatorname{cosec}^{2} 64^{\circ}$

$=3 \tan ^{2} 26^{\circ}-3 \operatorname{cosec}\left(90^{\circ}-26^{\circ}\right)$

$=3 \tan ^{2} 26^{\circ}-3 \sec ^{2} 26^{\circ}$

$=3\left(\tan ^{2} 26^{\circ}-\sec ^{2} 26^{\circ}\right)$

$=3 \times(-1)=-3 $  $\left\{\because \sec ^{2} \theta-\tan ^{2} \theta=1\right\}$

Ans (c)

Question 12

The value of $\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1$ is

(a) $-\cot \theta$

(b) $-\sin ^{2} \theta$

(c) $-\cos ^{2} \theta$

(d) $-\operatorname{cosec}^{2} \theta$

Sol :

$\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1$

$=\frac{\cos \theta \sin \theta}{\frac{\sin \theta}{\cos \theta}}-1$

$=\frac{\sin \theta \cos \theta \times \cos \theta}{\sin \theta}-1$

$=\cos ^{2} \theta-1=-\left(1-\cos ^{2} \theta\right)$

$=-\sin ^{2} \theta$

Ans (b)

ML Aggarwal Solution Class 10 Chapter 18 Trigonometric Identities Exercise 18

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31 Q32 Q33 Q34 Q35

 Exercise 18

Question 1

If A is an acute angle and sin A $=\frac{3}{5}$  find all other trigonometric ratios of angle A (using trigonometric identities).

Sol :

$\sin A=\frac{3}{5}$

In ∆ABC, ∠B = 90°

AC = 5 and BC = 3

Figure to be added

$\therefore \mathrm{AB}=\sqrt{\mathrm{AC}^{2}-\mathrm{BC}^{2}}=\sqrt{5^{2}-3^{2}}$

$=\sqrt{25-9}=\sqrt{16}=4$

Now, $\cos \theta=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4}{5}$

$\tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3}{4}$

$\cot \theta=\frac{1}{\tan \theta}=\frac{4}{3}$

$\sec \theta=\frac{1}{\cos \theta}=\frac{5}{4}$

$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{5}{3}$

Question 2

If A is an acute angle and sec $A=\frac{17}{8}$ , find all other trigonometric ratios of angle A (using trigonometric identities).

Sol :

$\sec A=\frac{17}{8}(A$ is an acute angle $)$

In right ∆ABC

$\sec A=\frac{A C}{A B}=\frac{17}{8}$

AC = 17, AB = 8

$\mathrm{BC}=\sqrt{\mathrm{AC}^{2}-\mathrm{AB}^{2}}=\sqrt{17^{2}-8^{2}}$

$=\sqrt{289-64}=\sqrt{225}=15$

Figure to be added

Now $\sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{15}{17}$

$\cos A=\frac{1}{\sec A}=\frac{8}{17}$

$\tan A=\frac{B C}{A B}=\frac{15}{8}$

$\cot A=\frac{1}{\tan A}=\frac{8}{15}$

$\operatorname{cosec} A=\frac{1}{\sin A}=\frac{17}{15}$

Question 3

Express the ratios cos A, tan A and sec A in terms of sin A.

Sol :

$\cos A=\sqrt{1-\sin ^{2} A}$

$\tan A=\frac{\sin A}{\operatorname{cos} A}=\frac{\sin A}{\sqrt{1-\sin ^{2} A}}$

$\sec A=\frac{1}{\cos A}=\frac{1}{\sqrt{1-\sin ^{2} A}}$

Question 4

If $\tan A=\frac{1}{\sqrt{3}}$, find all other trigonometric ratios of angle A.

Sol :

$\tan A=\frac{1}{\sqrt{3}}$

In right ∆ABC,

$\tan A=\frac{B C}{A B}=\frac{1}{\sqrt{3}}$

$\therefore B C=1, A B=\sqrt{3}$

$\mathrm{AC}=\sqrt{\mathrm{AB}^{2}+\mathrm{BC}^{2}}=\sqrt{(\sqrt{3})^{2}+(1)^{2}}$

$=\sqrt{3+1}=\sqrt{4}=2$

Figure to be added

$\therefore \sin A=\frac{B C}{A C}=\frac{1}{2}$

$\cos A=\frac{A B}{A C}=\frac{\sqrt{3}}{2}$

$\cot A=\frac{1}{\tan A}=\sqrt{3}$

$\sec A=\frac{1}{\cos A}=\frac{2}{\sqrt{3}}$

$\operatorname{cosec} A=\frac{1}{\sin A}=\frac{2}{1}=2$

Question 5

If 12 cosec θ = 13, find the value of $\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}$

Sol :

12 cosec θ = 13

$\Rightarrow \operatorname{cosec} \theta=\frac{13}{12}$

In right ∆ABC,

∠A = θ

$\operatorname{cosec} \theta=\frac{A C}{B C}=\frac{13}{12}$

Figure to be added

$\therefore A C=13, B C=12$

$A B=\sqrt{A C^{2}-B C^{2}}=\sqrt{13^{2}-12^{2}}$

$=\sqrt{169-144}=\sqrt{25}=5$

Now $\sin \theta=\frac{B C}{A C}=\frac{12}{13}$

$\cos \theta=\frac{A B}{A C}=\frac{5}{13}$

Now $\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}=\frac{2 \times \frac{12}{13}-3 \times \frac{5}{13}}{4 \times \frac{12}{13}-9 \times \frac{5}{13}}$

$=\frac{\frac{24}{13}-\frac{15}{13}}{\frac{48}{13}-\frac{45}{13}}$

$=\frac{\frac{9}{13}}{\frac{3}{13}}=\frac{9}{13} \times \frac{13}{3}=3$

Without using trigonometric tables, evaluate the following (6 to 10) :

Question 6

(i) $\cos ^{2} 26^{\circ}+\cos 64^{\circ} \sin 26^{\circ}+\frac{\tan 36^{\circ}}{\cot 54^{\circ}}$

(ii) $\frac{\sec 17^{\circ}}{\operatorname{cosec} 73^{\circ}}+\frac{\tan 68^{\circ}}{\cot 22^{\circ}}+\cos ^{2} 44^{\circ}+\cos ^{2} 46^{\circ}$

Sol :

Given that

(i) $\cos ^{2} 26^{\circ}+\cos 64^{\circ} \sin 26^{\circ}+\frac{\tan 36^{\circ}}{\cot 54^{\circ}}$

$\cos ^{2} 26^{\circ}+\cos \left(90^{\circ}-26^{\circ}\right) \sin 26^{\circ}+\frac{\tan 36^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}$

$=\cos ^{2} \cdot 26^{\circ}+\sin ^{2} 26^{\circ}+\frac{\tan 36^{\circ}}{\tan 36^{\circ}}$

=1+1=2

$\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right.$ and $\left.\cot \left(90^{\circ}-\theta\right)=\tan \theta, \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

(ii) $\frac{\sec 17^{\circ}}{\operatorname{cosec} 73^{\circ}}+\frac{\tan 68^{\circ}}{\cot 22^{\circ}}+\cos ^{2} 44^{\circ}+\cos ^{2} 46^{\circ}$

$\frac{\sec 17^{\circ}}{\operatorname{cosec} 73^{\circ}}+\frac{\tan 68^{\circ}}{\cot 22^{\circ}}+\cos ^{2} 44^{\circ}+\cos ^{2} 46^{\circ}$

$=\frac{\sec \left(90^{\circ}-73^{\circ}\right)}{\operatorname{cosec}\left(73^{\circ}\right)}+\frac{\tan \left(90^{\circ}-22^{\circ}\right)}{\cot 22^{\circ}}+\cos ^{2}\left(90^{\circ}-46^{\circ}+\cos ^{2} 46^{\circ}\right.$

$=\frac{\operatorname{cosec} 73^{\circ}}{\operatorname{cosec} 73^{\circ}}+\frac{\cot 22^{\circ}}{\cot 22^{\circ}}+\sin ^{2} 46^{\circ}+\cos ^{2} 46^{\circ}$ $\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

=1+1+1=3

Question 7

(i) $\frac{\sin 65^{\circ}}{\cos 25^{\circ}}+\frac{\cos 32^{\circ}}{\sin 58^{\circ}}-\sin 28^{\circ} \sec 62^{\circ}+\operatorname{cosec}^{2} 30^{\circ}(2015)$

(ii) $\frac{\sin 29^{\circ}}{\operatorname{cosec} 61^{\circ}}+2 \cot 8^{\circ} \cot 17^{\circ} \cot 45^{\circ} \cot 73^{\circ} \cot 82^{\circ}-3\left(\sin ^{2} 38^{\circ}+\sin ^{2} 52^{\circ}\right)$

Sol :

Given :

(i) $\frac{\sin 65^{\circ}}{\cos 25^{\circ}}+\frac{\cos 32^{\circ}}{\sin 58^{\circ}}-\sin 28^{\circ} \sec 62^{\circ}+\operatorname{cosec}^{2} 30^{\circ}$

$=\frac{\sin 65^{\circ}}{\cos \left(90^{\circ}-65^{\circ}\right)}+\frac{\cos 32^{\circ}}{\sin \left(90^{\circ}-32^{\circ}\right)}-\sin 28^{\circ} \times \sec \left(90^{\circ}-28^{\circ}\right)+(2)^{2}$

$=\frac{\sin 65^{\circ}}{\sin 65^{\circ}}+\frac{\cos 32^{\circ}}{\cos 32^{\circ}}-\sin 28^{\circ} \times \operatorname{cosec} 28^{\circ}+4$ $\left\{\begin{array}{l}\sin \left(90^{\circ}+\theta\right)=\cos \theta \\ \cos \left(90^{\circ}-\theta\right)=\sin \theta\end{array}\right\}$

=1+1-1+4

=6-1=5

(ii) $\frac{\sin 29^{\circ}}{\operatorname{cosec} 61^{\circ}}+2 \cot 8^{\circ} \cot 17^{\circ} \cot 45^{\circ} \cot 73^{\circ} \cot 82^{\circ}-3\left(\sin ^{2} 38^{\circ}+\sin ^{2} 52^{\circ}\right)$

$=\frac{\sec 29^{\circ}}{\operatorname{cosec}\left(90^{\circ}-29^{\circ}\right)}+2 \cot 8^{\circ} \cot 82^{\circ} \cot 17^{\circ} \cot 73^{\circ} \cot 45^{\circ}-3\left(\sin ^{2} 38^{\circ}+\sin ^{2}\left(90^{\circ}-38^{\circ}\right)\right.$

$\frac{\sec 29^{\circ}}{\sec 29^{\circ}}+2 \cot 8 \cot \left(90^{\circ}-8^{\circ}\right) \cot 17^{\circ} \cot \left(90^{\circ}-17^{\circ}\right) \cot 45^{\circ}-3\left(\sin ^{2} 38^{\circ}+\cos ^{2} 38^{\circ}\right)$

$=1+2 \cot 8^{\circ} \tan 8^{\circ} \cot 17^{\circ} \cot 17^{\circ} \cot 45^{\circ}-3(1)$

$\left\{\sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta, \cot \theta=\tan \left(90^{\circ}-\theta\right)\right.$ and $\left.\sin ^{2} \theta+\cos ^{2} \theta=1\right\}$

$=1+2 \times 1 \times 1 \times 1-3 \quad\left(\cot 45^{\circ}=1\right)$

=1+2-3=0

Question 8

(i) $\frac{\sin 35^{\circ} \cos 55^{\circ}+\cos 35^{\circ} \sin 55^{\circ}}{\operatorname{cosec}^{2} 10^{\circ}-\tan ^{2} 80^{\circ}}$

(ii) $\sin ^{2} 34^{\circ}+\sin ^{2} 56^{\circ}+2 \tan 18^{\circ} \tan 72^{\circ}-\cot ^{2} 30^{\circ}$

Sol :

Given :

(i) $\frac{\sin 35^{\circ} \cos 55^{\circ}+\cos 35^{\circ} \sin 55^{\circ}}{\operatorname{cosec}^{2} 10^{\circ}-\tan ^{2} 80^{\circ}}$

$=\frac{\sin 35^{\circ} \cos \left(90^{\circ}-35^{\circ}\right)+\cos 35^{\circ} \sin \left(90^{\circ}-35^{\circ}\right)}{\operatorname{cosec}^{2} 10^{\circ}-\tan ^{2}\left(90^{\circ}-10^{\circ}\right)}$

$=\frac{\sin 35^{\circ} \sin 35^{\circ}+\cos 35^{\circ} \cos 35^{\circ}}{\operatorname{cosec}^{2} 10^{\circ}-\cot ^{2} 10^{\circ}}$

$=\frac{\sin ^{2} 35^{\circ}+\cos ^{2} 35^{\circ}}{\operatorname{cosec}^{2} 10^{\circ}-\cot ^{2} 10^{\circ}}=\frac{1}{1}=1$ $\left\{\begin{array}{c}\because \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\end{array}\right\}$

(ii) $\sin ^{2} 34^{\circ}+\sin ^{2} 56^{\circ}+2 \tan 18^{\circ} \tan 72^{\circ}-\cot ^{2} 30^{\circ}$

$=\sin ^{2} 34^{\circ}+\sin ^{2}\left(90^{\circ}-34^{\circ}\right)+2 \tan 18^{\circ} \tan \left(90^{\circ}-18^{\circ}\right)-\cot ^{2} 30^{\circ}$

$=\sin ^{2} 34^{\circ}+\cos ^{2} 34^{\circ}+2 \tan 18^{\circ} \cot 18^{\circ}-\cot ^{2} 30^{\circ}$

$=1+2 \times 1-(\sqrt{3})^{2}$

=1+2-3=0

Question 9

(i) $\left(\frac{\tan 25^{\circ}}{\operatorname{cosec} 65^{\circ}}\right)^{2}+\left(\frac{\cot 25^{\circ}}{\sec 65^{\circ}}\right)^{2}+2 \tan 18^{\circ} \tan 45^{\circ} \tan 72^{\circ}$

(ii) $\left(\cos ^{2} 25+\cos ^{2} 65\right)+\operatorname{cosec} \theta \sec \left(90^{\circ}-\theta\right)-\cot \theta \tan \left(90^{\circ}-\theta\right)$

Sol :

(i) $\left(\frac{\tan 25^{\circ}}{\operatorname{cosec} 65^{\circ}}\right)^{2}+\left(\frac{\cot 25^{\circ}}{\sec 65^{\circ}}\right)^{2}+2 \tan 18^{\circ} \tan 45^{\circ} \tan 72^{\circ}$

$=\left(\frac{\tan 25^{\circ}}{\operatorname{cosec}\left(90^{\circ}-25^{\circ}\right)}\right)^{2}+\left(\frac{\cot 25^{\circ}}{\sec \left(90^{\circ}-25^{\circ}\right)}\right)^{2}+2 \tan 18^{\circ} \tan \left(90^{\circ}-18^{\circ}\right) \tan 45^{\circ}$

$=\left(\frac{\tan 25^{\circ}}{\sec 25^{\circ}}\right)^{2}+\left(\frac{\cot 25^{\circ}}{\operatorname{cosec} 25^{\circ}}\right)^{2}+2 \tan 18^{\circ} \cot 18^{\circ} \tan 45^{\circ}$

$=\left(\frac{\sin 25^{\circ} \times \cos 25^{\circ}}{\cos 25^{\circ} \times 1}\right)^{2}+\left(\frac{\cos 25^{\circ} \times \sin 25^{\circ}}{\sin 25^{\circ} \times 1}\right)+2 \times 1 \times 1$

$=\sin ^{2} 25^{\circ}+\cos ^{2} 25^{\circ}+2$

=1+2=3 $\left\{\begin{array}{c}\because \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \tan \theta \cot \theta=1\end{array}\right\}$

(ii) $\left(\cos ^{2} 25^{\circ}+\cos ^{2} 65^{\circ}\right)+\operatorname{cosec} \theta \sec \left(90^{\circ}-\theta\right)-\cot \theta \tan \left(90^{\circ}-\theta\right)$

$=\left[\cos ^{2} 25^{\circ}+\cos ^{2}\left(90^{\circ}-25^{\circ}\right)\right]+\operatorname{cosec} \theta \operatorname{cosec} \theta-\cot \theta \cdot \cot \theta$

$=\left(\cos ^{2} 25^{\circ}+\sin ^{2} 25^{\circ}\right)+\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)$

=1+1=2

Question 10

(i) $2\left(\sec ^{2} 35^{\circ}-\cot ^{2} 55^{\circ}\right)-\frac{\cos 28^{\circ} \operatorname{cosec} 62^{\circ}}{\tan 18^{\circ} \tan 36^{\circ} \tan 30^{\circ} \tan 54^{\circ} \tan 72^{\circ}}$

(ii) $\frac{\operatorname{cosec}^{2}(90-\theta)-\tan ^{2} \theta}{2\left(\cos ^{2} 48^{\circ}+\cos ^{2} 42^{\circ}\right)}-\frac{2 \tan ^{2} 30^{\circ} \sec ^{2} 52^{\circ} \sin ^{2} 38^{\circ}}{\operatorname{cosec}^{2} 70^{\circ}-\tan ^{2} 20^{\circ}}$

Sol :

(i) $2\left(\sec ^{2} 35^{\circ}-\cot ^{2} 55^{\circ}\right)-\frac{\cos 28^{\circ} \operatorname{cosec} 62^{\circ}}{\tan 18^{\circ} \tan 36^{\circ} \tan 30^{\circ} \tan 54^{\circ} \tan 72^{\circ}}$

$=2\left[\sec ^{2} 35^{\circ}-\cot ^{2}\left(90^{\circ}-35^{\circ}\right)\right]-\frac{\cos 28^{\circ} \operatorname{cosec}\left(90^{\circ}-28^{\circ}\right)}{\tan 18^{\circ} \tan \left(90^{\circ}-18^{\circ}\right) \tan 36^{\circ} \tan \left(90^{\circ}-36^{\circ}\right) \tan 30^{\circ}}$

$=2(1)-\frac{1}{1 \times 1 \times \frac{1}{\sqrt{3}}}=2-\frac{\sqrt{3}}{1}=2-\sqrt{3}$ $\left\{\begin{array}{c}\because \sec ^{2} \theta-\tan ^{2} \theta=1 \\ \tan \theta \cot \theta=1 \\ \cos \theta \sec \theta=1\end{array}\right\}$

(ii) $\frac{\operatorname{cosec}^{2}\left(90^{\circ}-\theta\right)-\tan ^{2} \theta}{2\left(\cos ^{2} 48^{\circ}+\cos ^{2} 42^{\circ}\right)}-\frac{2\left(\tan ^{2} 30^{\circ} \sec ^{2} 52^{\circ} \sin ^{2} 38^{\circ}\right)}{\operatorname{cosec}^{2} 70^{\circ}-\tan ^{2} 20^{\circ}}$

$=\frac{\sec ^{2} \theta-\tan ^{2} \theta}{2\left(\cos ^{2} 48^{\circ}+\cos ^{2}\left(90^{\circ}-48^{\circ}\right)\right.}-\frac{2\left[\tan ^{2} 30^{\circ} \sec ^{2} 52^{\circ} \sin ^{2}\left(90^{\circ}-52^{\circ}\right)\right.}{\operatorname{cosec}^{2} 70^{\circ}-\tan ^{2}\left(90^{\circ}-70^{\circ}\right)}$

$=\frac{1}{2\left[\cos ^{2} 48^{\circ}+\sin ^{2} 48^{\circ}\right]}-\frac{2\left[\left(\frac{1}{\sqrt{3}}\right)^{2} \sec ^{2} 52^{\circ} \cos ^{2} 52^{\circ}\right]}{\operatorname{cosec}^{2} 70^{\circ}-\cot ^{2} 70}$

$=\frac{1}{2 \times 1}-\frac{2\left[\frac{1}{3} \times 1\right]}{1}$

$=\frac{1}{2}-\frac{2}{3}$  $\left\{\begin{array}{c}\because \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \sec ^{2} \theta-\tan ^{2} \theta=1 \\ \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\end{array}\right\}$

$=\frac{3-4}{6}=\frac{-1}{6}$

Question 11

Prove that following:

(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1

(ii) $\frac{\tan \theta}{\tan \left(90^{\circ}-\theta\right)}+\frac{\sin \left(90^{\circ}-\theta\right)}{\cos \theta}=\sec ^{2} \theta$

(iii) $\frac{\cos \left(90^{\circ}-\theta\right) \cos \theta}{\tan \theta}+\cos ^{2}\left(90^{\circ}-\theta\right)=1$

(iv) $\sin \left(90^{\circ}-\theta\right) \cos \left(90^{\circ}-\theta\right)=\frac{\tan \theta}{1+\tan ^{2} \theta}$

Sol :

(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1

L.H.S. = cos θ sin (90° – θ) + sin θ cos (90° – θ)

= cos θ . cos θ + sin θ . sin θ

$=\cos ^{2} \theta+\sin ^{2} \theta$ = 1 = R.H.S.

(ii) $\frac{\tan \theta}{\tan \left(90^{\circ}-\theta\right)}+\frac{\sin \left(90^{\circ}-\theta\right)}{\cos \theta}=\sec ^{2} \theta$

L.H.S. $=\frac{\tan \theta}{\tan \left(90^{\circ}-\theta\right)}+\frac{\sin \left(90^{\circ}-\theta\right)}{\cos \theta}$

$=\frac{\tan \theta}{\cot \theta}+\frac{\cos \theta}{\cos \theta}=\tan \theta \times \tan \theta+1$

$=\tan ^{2} \theta+1=\sec ^{2} \theta=$R.H.S

(iii) $\frac{\cos \left(90^{\circ}-\theta\right) \cos \theta}{\tan \theta}+\cos ^{2}\left(90^{\circ}-\theta\right)=1$

L.H.S$=\frac{\cos \left(90^{\circ}-\theta\right) \cos \theta}{\tan \theta}+\cos ^{2}\left(90^{\circ}-\theta\right)$

$=\frac{\sin \theta \cos \theta}{\frac{\sin \theta}{\cos \theta}}+\sin ^{2} \theta$

$=\frac{\sin \theta \cos \theta \times \cos \theta}{\sin \theta}+\sin ^{2} \theta$

$=\cos ^{2} \theta+\sin ^{2} \theta=1$=R.H.S

(iv) $\sin \left(90^{\circ}-\theta\right) \cos \left(90^{\circ}-\theta\right)=\frac{\tan \theta}{1+\tan ^{2} \theta}$

L.H.S$=\sin \left(90^{\circ}-\theta\right) \cos \left(90^{\circ}-\theta\right)$

$=\cos \theta \sin \theta$ $\left\{\begin{array}{c}\because \sin \left(90^{\circ}-\theta\right)-\cos \theta \\ \sin ^{2} \theta+\cos ^{2} \theta=1\end{array}\right\}$

R.H.S$=\frac{\tan \theta}{1+\tan ^{2} \theta}=\frac{\frac{\sin \theta}{\cos \theta}}{1+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}$

$=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos ^{2} \theta}}$

$=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos ^{2} \theta}}=\frac{\sin \theta}{\cos \theta} \times \cos ^{2} \theta=\sin \theta \cos \theta$

∴L.H.S=R.H.S

Prove that following (12 to 30) identities, where the angles involved are acute angles for which the trigonometric ratios as defined:

Question 12

(i) (sec A + tan A) (1 – sin A) = cos A

(ii) (1 + $tan^2 A$) (1 – sin A) (1 + sin A) = 1.

Sol :

(i) (sec A + tan A) (1 – sin A) = cos A

L.H.S. = (sec A + tan A) (1 – sin A)

$=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)$

$=\frac{\cos ^{2} A}{\cos A}=\cos A$=R.H.S

$\left(1-\sin ^{2} A=\cos ^{2} A\right)$

(ii) $\left(1+\tan ^{2} A\right)(1-\sin A)(1+\sin A)=1$

L.H.S=$\left(1+\tan ^{2} A\right)(1-\sin A)(1+\sin A)$

$=\left(1+\frac{\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}\right)\left(1-\sin ^{2} \mathrm{~A}\right)$

$=\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A} \times \cos ^{2} A$

$\left\{\begin{array}{r}\because 1-\sin ^{2} A=\cos ^{2} A \\ \sin ^{2} A+\cos ^{2} A=1\end{array}\right\}$

$=\frac{1}{\cos ^{2} A} \times \cos ^{2} A=1$=R.H.S

Question 13

(i) tan A + cot A = sec A cosec A

(ii) (1 – cos A)(1 + sec A) = tan A sin A.

Sol :

(i) tan A + cot A = sec A cosec A

L.H.S. = tan A + cot A

$=\frac{\sin \mathrm{A}}{\cos \mathrm{A}}+\frac{\cos \mathrm{A}}{\sin \mathrm{A}}$

$=\frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}$

$=\frac{1}{\sin \mathrm{A} \cos \mathrm{A}}=\operatorname{cosec} \mathrm{A} \sec \mathrm{A}=\sec \mathrm{A} \operatorname{cosec}$

A=R.H.S

(ii) (1-cos A)(1+sec A)=tan A sin A

L.H.S=(1-cos A)(1+sec A)

$=(1-\cos A)\left(1+\frac{1}{\cos A}\right)$

$=(1-\cos A) \frac{(\cos A+1)}{\cos A}$

$=\frac{(1-\cos A)(1+\cos A)}{\cos A}=\frac{1-\cos ^{2} A}{\cos A}$

$=\frac{\sin ^{2} A}{\cos A}$

$=\frac{\sin ^{2} A}{\cos A}=\sin A \times \frac{\sin A}{\cos A}$ $\left\{1-\cos ^{2} A=\sin ^{2} A\right\}$

=tan A sin A =R.H.S

Question 14

(i) $\frac{1}{1+\cos A}+\frac{1}{1-\cos A}=2 \operatorname{cosec}^{2} A$

(ii) $\frac{1}{\sec A+\tan A}+\frac{1}{\sec A-\tan A}=2 \sec A$

Sol :

(i) $\frac{1}{1+\cos A}+\frac{1}{1-\cos A}=2 \operatorname{cosec}^{2} A$

L.H.S$=\frac{1}{1+\cos A}+\frac{1}{1-\cos A}$

$=\frac{1-\cos A+1+\cos A}{(1+\cos A)(1-\cos A)}$ 

$=\frac{2}{1-\cos ^{2} A}=\frac{2}{\sin ^{2} A}$ $\left(\because 1-\cos ^{2} A=\sin ^{2} A\right)$

$=2 \operatorname{cosec}^{2} A$=R.H.S

(ii) $\frac{1}{\sec A+\tan A}+\frac{1}{\sec A-\tan A}=2 \sec A$

L.H.S $=\frac{1}{\sec A+\tan A}+\frac{1}{\sec A-\tan A}$

$=\frac{\sec A-\tan A+\sec A+\tan A}{(\sec A+\tan A)(\sec A-\tan A)}$

$=\frac{2 \sec A}{\sec ^{2} A-\tan ^{2} A}=\frac{2 \sec A}{1}$ $\left(\because \sec ^{2} A-\tan ^{2} A=1\right)$

=2 sec A =R.H.S

Question 15

(i) $\frac{\sin A}{1+\cos A}=\frac{1-\cos A}{\sin A}$

(ii) $\frac{1-\tan ^{2} A}{\cot ^{2} A-1}=\tan ^{2} A$

(iii) $\frac{\sin A}{1+\cos A}=\operatorname{cosec} A-\cot A$

Sol :

(i) $\frac{\sin A}{1+\cos A}=\frac{1-\cos A}{\sin A}$

L.H.S$=\frac{\sin A}{1+\cos A}$

[multiplying and dividing by (1 – cosA)]

$=\frac{\sin A(1-\cos A)}{1-\cos ^{2} A}=\frac{\sin A(1-\cos A)}{\sin ^{2} A}$

$\left(\because 1-\cos ^{2} A=\sin ^{2} A\right)$

$=\frac{1-\cos A}{\sin A}$=R.H.S

(ii) $\frac{1-\tan ^{2} A}{\cot ^{2} A-1}=\tan ^{2} A$

L.H.S$=\frac{1-\tan ^{2} A}{\cot ^{2} A-1}$

$=\frac{1-\frac{\sin ^{2} A}{\cos ^{2} A}}{\frac{\cos ^{2} A}{\sin ^{2} A}-1}$

$=\frac{\frac{\cos ^{2} A-\sin ^{2} A}{\cos ^{2} A}}{\frac{\cos ^{2} A-\sin ^{2} A}{\sin ^{2} A}}$

$=\frac{\cos ^{2} A-\sin ^{2} A}{\cos ^{2} A} \times \frac{\sin ^{2} A}{\cos ^{2} A-\sin ^{2} A}$

$=\frac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A$= R.H.S

(iii) $\frac{\sin A}{1+\cos A}=\operatorname{cosec} A-\cot A$

R.H.S=cosec A- cot A

$=\frac{1}{\sin A}-\frac{\cos A}{\sin A}=\frac{1-\cos A}{\sin A}$

$=\frac{(1-\cos A)(1+\cos A)}{\sin A(1+\cos A)}$

( Multiplying and dividing by 1+cos A)

$=\frac{1-\cos ^{2} A}{\sin A(1+\cos A)}=\frac{\sin ^{2} A}{\sin A(1+\cos A)}$

$\left\{\because 1-\cos ^{2} A=\sin ^{2} A\right\}$

$\frac{\sin A}{1+\cos A}$=L.H .S

Question 16

(i) $\frac{\sec A-1}{\sec A+1}=\frac{1-\cos A}{1+\cos A}$

(ii) $\frac{\tan ^{2} \theta}{(\sec \theta-1)^{2}}=\frac{1+\cos \theta}{1-\cos \theta}$

(iii) $(1+\tan A)^{2}+(1-\tan A)^{2}=2 \sec ^{2} A$

(iv) $\sec ^{2} A+\operatorname{cosec}^{2} A=\sec ^{2} A \cdot \operatorname{cosec}^{2} A$

Sol :

(i) $\frac{\sec A-1}{\sec A+1}=\frac{1-\cos A}{1+\cos A}$

L.H.S$=\frac{\sec A-1}{\sec A+1}$

$=\frac{\frac{1-\cos A}{\cos A}}{\frac{1+\cos A}{\cos A}}=\frac{1-\cos A}{\cos A} \times \frac{\cos A}{1+\cos A}$

$=\frac{1-\cos A}{1+\cos A}=$R.H.S

(ii) Prove that $\frac{\tan ^{2} \theta}{(\sec \theta-1)^{2}}=\frac{1+\cos \theta}{1-\cos \theta}$

L.H.S$=\frac{\tan ^{2} \theta}{(\sec \theta-1)^{2}}=\frac{\tan ^{2} \theta}{\sec ^{2} \theta+1-2 \sec \theta}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{1}{\cos ^{2} \theta}+1-\frac{2}{\cos \theta}}$

$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{1+\cos ^{2} \theta-2 \cos \theta}$

$=\frac{\sin ^{2} \theta}{(1-\cos \theta)^{2}}$

$=\frac{\left(1-\cos ^{2} \theta\right)}{(1-\cos \theta)^{2}}$

$=\frac{(1+\cos \theta)(1-\cos \theta)}{(1-\cos \theta)^{2}}$

$=\frac{1+\cos \theta}{1-\cos \theta}$

=R.H.S

(ii) $(1+\tan A)^{2}+(1-\tan A)^{2}=2 \sec ^{2} A$

L.H.S $=(1+\tan A)^{2}+(1-\tan A)^{2}$

$=1+2 \tan A+\tan ^{2} A+1-2 \tan A+\tan ^{2} A$

$=2+2 \tan ^{2} A=2\left(1+\tan ^{2} A\right)$

$=2 \sec ^{2} A $ $\left(\because 1+\tan ^{2} A=\sec ^{2} A\right)$

=R.H.S

(iv) $\sec ^{2} A+\operatorname{cosec}^{2} A=\sec ^{2} A \operatorname{cosec}^{2} A$

L.H.S$=\sec ^{2} A+\operatorname{cosec}^{2} A$

$=\frac{1}{\cos ^{2} A}+\frac{1}{\sin ^{2} A}$

$=\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A+\cos ^{2} A}$

$=\frac{1}{\sin ^{2} A \cos ^{2} A}=\sec ^{2} A \operatorname{cosec}^{2} A$

=R.H.S

Question 17

(i) $\frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=2 \sec A$

(ii) $\frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=2 \operatorname{cosec} A$

Sol :

(i) $\frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=2 \sec A$

L.H.S$=\frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}$

$=\frac{(1+\sin A)(1+\sin A)+\cos ^{2} A}{\cos A(1+\sin A)}$

$=\frac{1+\sin A+\sin A+\sin ^{2} A+\cos ^{2} A}{\cos A(1+\sin A)}$

$=\frac{1+2 \sin \mathrm{A}+1}{\cos \mathrm{A}(1+\sin \mathrm{A})}=\frac{2+2 \sin \mathrm{A}}{\cos \mathrm{A}(1+\sin \mathrm{A})}$

$=\frac{2(1+\sin A)}{\cos A(1+\sin A)}=\frac{2}{\cos A}=2 \sec A$

=R.H.S

(ii) $\frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=2 \operatorname{cosec} A$

L.H.S$=\frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}$

$=\tan A\left(\frac{1}{\sec A-1}+\frac{1}{\sec A+1}\right)$

$=\tan A\left(\frac{\sec A+1+\sec A-1}{(\sec A-1)(\sec A+1)}\right)$

$=\frac{\tan A \times 2 \sec A}{\sec ^{2} A-1}$

$=\frac{2 \sec A \tan A}{\tan ^{2} A}$

$=\frac{2 \sec A}{\tan A}$

$=\frac{2 \times 1 \times \cos A}{\cos A \times \sin A}$

$=\frac{2}{\sin A}=2 \operatorname{cosec} A$

=R.H.S

Question 18

(i) $\frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1}=2 \sec ^{2} A$

(ii) $\cot A-\tan A=\frac{2 \cos ^{2} A-1}{\sin A-\cos A}$

(iii) $\frac{\cot A-1}{2-\sec ^{2} A}=\frac{\cot A}{1+\tan A}$

Sol :

(i) $\frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1}=2 \sec ^{2} A$

L.H.S$=\frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1}$

$=\operatorname{cosec} \mathrm{A}\left[\frac{1}{\operatorname{cosec} \mathrm{A}-1}+\frac{1}{\operatorname{cosec} \mathrm{A}+1}\right]$

$=\operatorname{cosec} A\left[\frac{\operatorname{cosec} A+1+\operatorname{cosec} A-1}{(\operatorname{cosec} A-1)(\operatorname{cosec} A+1)}\right]$

$=\frac{\operatorname{cosec} A \times 2 \operatorname{cosec} A}{\operatorname{cosec}^{2} A-1}=\frac{2 \operatorname{cosec}^{2} A}{\cot ^{2} A}$

$=\frac{2 \times \sin ^{2} A}{\sin ^{2} A \times \cos ^{2} A}=\frac{2}{\cos ^{2} A}$

$=2 \sec ^{2} A$=R.H.S

(ii) $\cot A-\tan A=\frac{2 \cos ^{2} A-1}{\sin A-\cos A}$

L.H.S=cot A-tan A

$=\frac{\cos A}{\sin A}-\frac{\sin A}{\cos A}=\frac{\cos ^{2} A-\sin ^{2} A}{\sin A \cos A}$

$=\frac{\cos ^{2} A-\left(1-\cos ^{2} A\right)}{\sin A \cos A}$

$=\frac{\cos ^{2} A-1+\cos ^{2} A}{\sin A \cos A}$

$=\frac{2 \cos ^{2} A-1}{\sin A \cos A}$

=R.H.S

(iii) $\frac{\cot A-1}{2-\sec ^{2} A}=\frac{\cot A}{1+\tan A}$

L.H.S$=\frac{\cot A-1}{2-\sec ^{2} A}$

$=\frac{\frac{\cos A}{\sin A}-1}{2-\frac{1}{\cos ^{2} A}}$

$=\frac{\frac{\cos A-\sin A}{\sin A}}{\frac{2 \cos ^{2} A-1}{\cos ^{2} A}}$

$=\frac{\cos A-\sin A}{\sin A} \times \frac{\cos ^{2} A}{2 \cos ^{2} A-1}$

$=\frac{\cos ^{2} A(\cos A-\sin A)}{\sin A\left(2 \cos ^{2} A-1\right)}$

$=\frac{\cos ^{2} A(\cos A-\sin A)}{\sin A\left[2 \cos ^{2} A-\left(\sin ^{2} A+\cos ^{2} A\right)\right]}$

$=\frac{\cos ^{2} A(\cos A-\sin A)}{\sin A\left[2 \cos ^{2} A-\sin ^{2} A-\cos ^{2} A\right]}$

$=\frac{\cos ^{2} A(\cos A-\sin A)}{\sin A\left(\cos ^{2} A-\sin ^{2} A\right)}$

$=\frac{\cos ^{2} A(\cos A-\sin A)}{\sin A(\cos A+\sin A)(\cos A-\sin A)}$

$=\frac{\cos ^{2} A}{\sin A(\cos A+\sin A)}$

R.H.S$=\frac{\cot A}{1+\tan A}=\frac{\frac{\cos A}{\sin A}}{1+\frac{\sin A}{\cos A}}$

$=\frac{\frac{\cos A}{\sin A}}{\frac{\cos A+\sin A}{\cos A}}$

$=\frac{\cos A}{\sin A} \times \frac{\cos A}{\cos A+\sin A}$

$=\frac{\cos ^{2} A}{\sin A(\cos A+\sin A)}$

∴L.H.S=R.H.S

Question 19

(i) $\tan ^{2} \theta-\sin ^{2} \theta=\tan ^{2} \theta \sin ^{2} \theta$

(ii) $\frac{\cos \theta}{1-\tan \theta}-\frac{\sin ^{2} \theta}{\cos \theta-\sin \theta}=\cos \theta+\sin \theta$

Sol :

(i) $\tan ^{2} \theta-\sin ^{2} \theta=\tan ^{2} \theta \sin ^{2} \theta$

L.H.S $=\tan ^{2} \theta-\sin ^{2} \theta$

$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}-\sin ^{2} \theta$

$=\frac{\sin ^{2} \theta-\sin ^{2} \theta \cos ^{2} \theta}{\cos ^{2} \theta}$

$=\frac{\sin ^{2} \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}=\sin ^{2} \theta \times \frac{\sin ^{2} \theta}{\cos ^{2} \theta}$

$=\sin ^{2} \theta \times \tan ^{2} \theta$

$=\tan ^{2} \theta \sin ^{2} \theta$

=R.H.S

(ii) $\frac{\cos \theta}{1-\tan \theta}-\frac{\sin ^{2} \theta}{\cos \theta-\sin \theta}=\cos \theta+\sin \theta$

L.H.S$=\frac{\cos \theta}{1-\tan \theta}-\frac{\sin ^{2} \theta}{\cos \theta-\sin \theta}$

$=\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}-\frac{\sin ^{2} \theta}{\cos \theta-\sin \theta}$

$=\frac{\cos \theta}{\frac{\cos \theta-\sin \theta}{\cos \theta}}-\frac{\sin ^{2} \theta}{\cos \theta-\sin \theta}$

$=\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}-\frac{\sin ^{2} \theta}{\cos \theta-\sin \theta}$

$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos \theta-\sin \theta}$

$=\frac{(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)}{\cos \theta-\sin \theta}$

=cos θ+sin θ=R.H.S

Question 20

(i) $\operatorname{cosec}^{4} \theta-\operatorname{cosec}^{2} \theta=\cot ^{4} \theta+\cot ^{2} \theta$

(ii) $2 \sec ^{2} \theta-\sec ^{4} \theta-2 \operatorname{cosec}^{2} \theta+\operatorname{cosec}^{4} \theta=\cot ^{4} \theta-\tan ^{4} \theta$

Sol :

(i) $\operatorname{cosec}^{4} \theta-\operatorname{cosec}^{2} \theta=\cot ^{4} \theta+\cot ^{2} \theta$

L.H .S$=\operatorname{cosec}^{4} \theta-\operatorname{cosec}^{2} \theta$

$=\operatorname{cosec}^{2} \theta\left(\operatorname{cosec}^{2} \theta-1\right)$

$=\operatorname{cosec}^{2} \theta \cot ^{2} \theta$ $\left(\operatorname{cosec}^{2} \theta-1=\cot ^{2} \theta\right)$

$=\left(\cot ^{2} \theta+1\right) \cot ^{2} \theta$

$=\cot ^{4} \theta+\cot ^{2} \theta$

=R.H.S

(ii) $2 \sec ^{2} \theta-\sec ^{4} \theta-2 \operatorname{cosec}^{2} \theta+\operatorname{cosec}^{4} \theta=\cot ^{4} \theta-\tan ^{4} \theta$

L.H.S$=2 \sec ^{2} \theta-\sec ^{4} \theta-2 \operatorname{cosec}^{2} \theta+\operatorname{cosec}^{4} \theta$

$=2\left(\tan ^{2} \theta+1\right)-\left(\tan ^{2} \theta+1\right)^{2}-2(1+\left.\cot ^{2} \theta\right)+\left(1+\cot ^{2} \theta\right)^{2}$

$\left\{\begin{array}{r}\because \sec ^{2} \theta=\tan ^{2} \theta+1 \\ \operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta\end{array}\right\}$

$=2 \tan ^{2} \theta+2-\left(\tan ^{4} \theta+2 \tan ^{2} \theta+1\right)-2-2 \cot ^{2} \theta+\left(1+2 \cot ^{2} \theta+\cot ^{4} \theta\right)$

$=2 \tan ^{2} \theta+2-\tan ^{4} \theta-2 \tan ^{2} \theta-1-2-2 \cot ^{2} \theta+1+2 \cot ^{2} \theta+\cot ^{4} \theta$

$=\cot ^{4} \theta-\tan ^{4} \theta$=R.H.S

Question 21

(i) $\frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta$

(ii) $\frac{\tan ^{3} \theta-1}{\tan \theta-1}=\sec ^{2} \theta+\tan \theta$

Sol :

(i) $\frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta$

L.H.S$=\frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}$

L.H.S. $=\frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}$

$=\frac{\cos \theta+\cos ^{2} \theta}{\sin \theta(1+\cos \theta)}$

$=\frac{\cos \theta(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$

$=\frac{\cos \theta}{\sin \theta}=\cot \theta$

=R.H.S

(iii) $\frac{\tan ^{3} \theta-1}{\tan \theta-1}=\sec ^{2} \theta+\tan \theta$

L.H.S. $=\frac{(\tan \theta-1)}{\tan \theta-1}\left(\tan ^{2} \theta+\tan \theta+1\right)$

$=\tan ^{2} \theta+\tan \theta+1=\tan ^{2} \theta+1+\tan \theta$

$=\sec ^{2} \theta+\tan \theta$

=R.H.S

Question 22

(i) $\frac{1+\operatorname{cosec} A}{\operatorname{cosec} A}=\frac{\cos ^{2} A}{1-\sin A}$

(ii) $\sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{\sin A}{1+\cos A}$

Sol :

(i) $\frac{1+\operatorname{cosec} A}{\operatorname{cosec} A}=\frac{\cos ^{2} A}{1-\sin A}$

$\mathrm{L.H.S}=\frac{1+\operatorname{cosec} A}{\operatorname{cosec} A}$

$=\frac{1+\operatorname{cosec} A}{\operatorname{cosec} A}$

$=\frac{1+\frac{1}{\sin A}}{\frac{1}{\sin A}}$

$=\frac{\sin A+1}{\sin A} \times \frac{\sin A}{1}$

=sin A+1

R.H.S$=\frac{\cos ^{2} A}{1-\sin A}=\frac{1-\sin ^{2} A}{1-\sin A}$

$=\frac{(1+\sin \mathrm{A})(1-\sin \mathrm{A})}{1-\sin \mathrm{A}}$

=1+sin A=sin A+1

∴L.H.S=R.H.S

(ii) $\sqrt{\frac{1-\cos A}{1+\cos A}}=\operatorname{cosec} A-\cot A$

$\mathrm{L.H.S.}=\sqrt{\frac{1-\cos \mathrm{A}}{1+\cos \mathrm{A}}}$

Rationalising the denominator

$=\sqrt{\frac{(1-\cos A)(1-\cos A)}{(1+\cos A)(1-\cos A)}}$

$=\sqrt{\frac{(1-\cos A)^{2}}{1-\cos ^{2} A}}$

$=\sqrt{\frac{(1-\cos A)^{2}}{\sin ^{2} A}}$

$=\frac{1-\cos A}{\sin A}$

$=\frac{1}{\sin A}-\frac{\cos A}{\sin A}$

=cosec A-cot A=R.H.S

Question 23

(i) $\sqrt{\frac{1+\sin A}{1-\sin A}}=\tan A+\sec A$

(ii) $\sqrt{\frac{1-\cos A}{1+\cos A}}=\operatorname{cosec} A-\cot A$

Sol :

(i) $\sqrt{\frac{1+\sin A}{1-\sin A}}=\tan A+\sec A$

L.H.S. $=\sqrt{\frac{1+\sin \mathrm{A}}{1-\sin \mathrm{A}}}$

Rationalising the denominator

$=\sqrt{\frac{(1+\sin \mathrm{A})(1+\sin \mathrm{A})}{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}}$

$=\sqrt{\frac{(1+\sin \mathrm{A})^{2}}{1-\sin ^{2} \mathrm{~A}}}$

$=\sqrt{\frac{(1+\sin \mathrm{A})^{2}}{\cos ^{2} \mathrm{~A}}}$

$=\frac{1+\sin A}{\cos A}$

$=\frac{1}{\cos A}+\frac{\sin A}{\cos A}$

=sec A+ tan A

=tan A+ sec A=R.H.S

(ii) $\sqrt{\frac{1-\cos A}{1+\cos A}}=\operatorname{cosec} A-\cot A$

L.H.S. $=\sqrt{\frac{1-\cos A}{1+\cos A}}$

Rationalising the denominator

$=\sqrt{\frac{(1-\cos A)(1-\cos A)}{(1+\cos A)(1-\cos A)}}$

$=\sqrt{\frac{(1-\cos A)^{2}}{1-\cos ^{2} A}}$

$=\sqrt{\frac{(1-\cos A)^{2}}{\sin ^{2} A}}$

$=\frac{1-\cos A}{\sin A}$

$=\frac{1}{\sin A}-\frac{\cos A}{\sin A}$

=cosec A-cot A=R.H.S

Question 24

(i) $\sqrt{\frac{\sec A-1}{\sec A+1}}+\sqrt{\frac{\sec A+1}{\sec A-1}}=2 \operatorname{cosec} A$

(ii) $\frac{\cot A \cot A}{1-\sin A}=1+\operatorname{cosec} A$

Sol :

(i) $\sqrt{\frac{\sec A-1}{\sec A+1}}+\sqrt{\frac{\sec A+1}{\sec A-1}}=2 \operatorname{cosec} A$

$\mathrm{LH.S}=\sqrt{\frac{\sec A-1}{\sec A+1}}+\sqrt{\frac{\sec A+1}{\sec A-1}}$

$=\frac{\sqrt{\sec A-1}}{\sqrt{\sec A+1}}+\frac{\sqrt{\sec A+1}}{\sqrt{\sec A-1}}$

$=\frac{\sec A-1+\sec A+1}{\sqrt{(\sec A+1)(\sec A-1)}}$

$=\frac{2 \sec A}{\sqrt{\sec ^{2} A-1}}$

$\left\{\because \sec ^{2} A-1=\tan ^{2} A\right\}$

$=\frac{2 \sec A}{\sqrt{\tan ^{2} A}}=\frac{2 \sec A}{\tan A}$

$=\frac{2 \times \cos A}{\cos A \times \sin A}=\frac{2}{\sin A}$

$=2 \operatorname{cosec} A$=R.H.S .

(ii) $\frac{\cos A \cot A}{1-\sin A}=1+\operatorname{cosec} A$

L.H.S $=\frac{\cos A \cot A}{1-\sin A}=\frac{\cos A \cos A}{\sin A(1-\sin A)}$

$\left\{\cos A=\frac{\cos A}{\sin A}\right\}$

$=\frac{\cos ^{2} A}{\sin A(1-\sin A)}=\frac{1-\sin ^{2} A}{\sin A(1-\sin A)}$

$\left\{\because \cos ^{2} A=1-\sin ^{2} A\right\}$

$=\frac{(1+\sin A)(1-\sin A)}{\sin A(1-\sin A)}=\frac{1+\sin A}{\sin A}$

$=\frac{1}{\sin A}+\frac{\sin A}{\sin A}=\operatorname{cosec} A+1$

=1+cosec A=R.H.S

Question 25

(i) $\frac{1+\tan A}{\sin A}+\frac{1+\cot A}{\cos A}=2(\sec A+\operatorname{cosec} A)$

(ii) $\sec ^{4} A-\tan ^{4} A=1+2 \tan ^{2} A$

Sol :

(i) $\frac{1+\tan A}{\sin A}+\frac{1+\cot A}{\cos A}=2(\sec A+\operatorname{cosec} A)$

L.H.S$=\frac{1+\tan A}{\sin A}+\frac{1+\cot A}{\cos A}$

$=\frac{1+\frac{\sin A}{\cos A}}{\sin A}+\frac{1+\frac{\cos A}{\sin A}}{\cos A}$

$=\frac{\cos A+\sin A}{\cos A \times \sin A}+\frac{\sin A+\cos A}{\cos A \times \sin A}$

$=2\left[\frac{\cos A+\sin A}{\cos A \sin A}\right]$

$=2\left[\frac{\cos A}{\cos A \sin A}+\frac{\sin A}{\cos A \sin A}\right]$

$=2\left[\frac{1}{\sin A}+\frac{1}{\cos A}\right]$

=2(cosec A+ sec A)

=2(sec A+cosec A)

=R.H.S

(ii) $\sec ^{4} A-\tan ^{4} A=1+2 \tan ^{2} A$

L.H.S$=\sec ^{4} A-\tan ^{4} A$

$=\left(\sec ^{2} A-\tan ^{2} A\right)\left(\sec ^{2} A+\tan ^{2} A\right)$

$=\left(1+\tan ^{2} A-\tan ^{2} A\right)\left(1+\tan ^{2} A+\tan ^{2} A\right)$

$\left\{\because \sec ^{2} A=\tan ^{2} A+1\right\}$

$=1\left(1+2 \tan ^{2} A\right)=1+2 \tan ^{2} A$=R.H.S

Question 26

(i) $\operatorname{cosec}^{6} A-\cot ^{6} A=3 \cot ^{2} A \operatorname{cosec}^{2} A+1$

(ii) $\sec ^{6} A-\tan ^{6} A=1+3 \tan ^{2} A+3 \tan ^{4} A$

Sol :

(i) $\operatorname{cosec}^{6} A-\cot ^{6} A=3 \cot ^{2} A \operatorname{cosec}^{2} A+1$

L.H.S$=\operatorname{cosec}^{6} A-\cot ^{6} A$

$=\left(\operatorname{cosec}^{2} A\right)^{3}-\left(\cot ^{2} A\right)^{3}$

$=\left(\operatorname{cosec}^{2} \theta-\cot ^{2} A\right)^{3}+3 \operatorname{cosec}^{2} A \cot ^{2} A\left(\operatorname{cosec}^{2} A-\cot ^{2} A\right)$

$=(1)^{3}+3 \operatorname{cosec}^{2} A \cot ^{2} A \times 1$

$=1+3 \cot ^{2} A \operatorname{cosec}^{2} A$

$=3 \cot ^{2} A \operatorname{cosec}^{2} A+1$=R.H.S

(ii) $\sec ^{6} A-\tan ^{6} A=1+3 \tan ^{2} A+3 \tan ^{4} A$

L.H.S$=\sec ^{6} A-\tan ^{6} A$

$=\left(\sec ^{2} A\right)^{3}-\left(\tan ^{2} A\right)^{3}$

$=\left(\sec ^{2} A-\tan ^{2} A\right)^{3}+3 \sec ^{2} A \tan ^{2} A\left(\sec ^{2} A-\tan ^{2} A\right)$

$=(1)^{3}+3 \sec ^{2} \mathrm{~A} \tan ^{2} \mathrm{~A} \times 1$

$=1+3 \sec ^{2} \mathrm{~A} \tan ^{2} \mathrm{~A}$

$=1+3\left[\left(1+\tan ^{2} A\right)\left(\tan ^{2} A\right)\right]$

$=1+3\left[\tan ^{2} A+\tan ^{4} A\right]$

$=1+3 \tan ^{2} A+3 \tan ^{4} A$=R.H.S

Question 27

(i) $\frac{\cot \theta-\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}=\frac{1+\cos \theta}{\sin \theta}$

(ii) $\frac{\sin \theta}{\cot \theta+\operatorname{cosec} \theta}=2+\frac{\sin \theta}{\cot \theta-\operatorname{cosec} \theta}$

Sol :

(i) $\frac{\cot \theta-\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}=\frac{1+\cos \theta}{\sin \theta}$

L.H.S $=\frac{\cot \theta-\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{\frac{\cos \theta}{\sin \theta}+\frac{1}{\sin \theta}-1}{\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}+1}$

$=\frac{\cos \theta+1-\sin \theta}{\sin \theta} \times \frac{\sin \theta}{\cos \theta-1+\sin \theta}$

$=\frac{\cos \theta+1-\sin \theta}{\cos \theta-1+\sin \theta}$

$=\frac{\cos \theta+(1-\sin \theta)}{\cos \theta-(1-\sin \theta)}$

$=\frac{[\cos \theta+(1-\sin \theta)][\cos \theta+(1-\sin \theta)]}{[\cos \theta-(1-\sin \theta)][\cos \theta+(1-\sin \theta)]}$

$=\frac{[\cos \theta+(1-\sin \theta)]^{2}}{\cos ^{2} \theta-(1-\sin \theta)^{2}}$

$=\frac{(\cos \theta+1-\sin \theta)^{2}}{\cos ^{2} \theta-\left(1+\sin ^{2} \theta-2 \sin \theta\right)}$

$=\frac{\cos ^{2} \theta+\sin ^{2} \theta+1+2 \cos \theta-2 \sin \theta-2 \sin \theta \cos \theta}{\cos ^{2} \theta-1-\sin ^{2} \theta+2 \sin \theta}$

$=\frac{1+1+2 \cos \theta-2 \sin \theta-2 \sin \theta \cos \theta}{1-\sin ^{2} \theta-1-\sin ^{2} \theta+2 \sin \theta}$

$=\frac{2+2 \cos \theta-2 \sin \theta-2 \sin \theta \cos \theta}{2 \sin \theta-2 \sin ^{2} \theta}$

$=\frac{2(1+\cos \theta)-2 \sin \theta(1+\cos \theta)}{2 \sin \theta(1-\sin \theta)}$

$=\frac{(1+\cos \theta) 2(1-\sin \theta)}{2 \sin \theta(1-\sin \theta)}$

$=\frac{1+\cos \theta}{\sin \theta}$=R.H.S

(ii) $\frac{\sin \theta}{\cot \theta+\operatorname{cosec} \theta}=2+\frac{\sin \theta}{\cot \theta-\operatorname{cosec} \theta}$

$=\frac{\sin ^{2} \theta}{1+\cos \theta}$

$=\frac{1-\cos ^{2} \theta}{1+\cos \theta}=\frac{(1+\cos \theta)(1-\cos \theta)}{1+\cos \theta}$

=1-cos θ

R.H.S$=2+\frac{\sin \theta}{\cot \theta-\operatorname{cosec} \theta}$

$=2+\frac{\sin \theta}{\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}}$

$=2+\frac{\sin \theta}{\frac{\cos \theta-1}{\sin \theta}}$

$=2+\frac{\sin ^{2} \theta}{\cos \theta-1}$

$=\frac{2 \cos \theta-2+\sin ^{2} \theta}{\cos \theta-1}$

$=\frac{2 \cos \theta-2+\left(1-\cos ^{2} \theta\right)}{\cos \theta-1}$

$=\frac{2(\cos \theta-1)+(1+\cos \theta)(1-\cos \theta)}{\cos \theta-1}$

Question 28

(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ

(ii) (cosecA – sinA)(secA – cosA) $sec^2A$ = tanA

(iii) (cosecθ – sinθ)(secθ – cosθ)(tan θ + cotθ) = 1

Sol :

(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ

L.H.S = (sinθ + cosθ)(secθ + cosecθ)

$=(\sin \theta+\cos \theta)\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right)$

$=\frac{(\sin \theta+\cos \theta)(\sin \theta+\cos \theta)}{\sin \theta \cos \theta}$

$=\frac{\sin ^{2} \theta+\sin \theta \cos \theta+\sin \theta \cos \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}$

$=\frac{1+2 \sin \theta \cos \theta}{\sin \theta \cos \theta}$

$=\frac{1}{\sin \theta \cos \theta}+\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}$

=cosec ፀ sec ፀ +2

=2+secθ cosecθ

(ii) (cosec A-sin A)(sec A-cos A)$\sec ^{2} A$

=tan A

L.H.S

$(\operatorname{cosec} A-\sin A)(\sec A-\cos A) \sec ^{2} A$

$=\left(\frac{1}{\sin \mathrm{A}}-\sin \mathrm{A}\right)\left(\frac{\mathrm{I}}{\cos \mathrm{A}}-\cos \mathrm{A}\right) \frac{1}{\cos ^{2} \mathrm{~A}}$

$=\left(\frac{1-\sin ^{2} A}{\sin A}\right)\left(\frac{1-\cos ^{2} A}{\cos A}\right) \frac{1}{\cos ^{2} A}$

$\frac{\cos ^{2} A}{\sin A} \cdot \frac{\sin ^{2} A}{\cos A} \cdot \frac{1}{\cos ^{2} A}=\frac{\sin A}{\cos A}=\tan A$

R.H.S

(iii) (cosecθ -sinθ)(secθ-cosθ)(tanθ+cotθ)=1

L.H.S=(cosecθ-sinθ)(secθ-cosθ)(tanθ+cotθ)

$=\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta\right)$(tanθ+cotθ)

$=\frac{1-\sin ^{2} \theta}{\sin \theta} \times \frac{1-\cos ^{2} \theta}{\cos \theta}(\tan \theta+\cot \theta)$

$=\frac{\cos ^{2} \theta}{\sin \theta} \times \frac{\sin ^{2} \theta}{\cos \theta}(\tan \theta+\cot \theta)$

$=\sin \theta \cos \theta \frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}$

=1=R.H.S

Question 29

(i) $\frac{\sin ^{3} A+\cos ^{3} A}{\sin A+\cos A}+\frac{\sin ^{3} A-\cos ^{3} A}{\sin A-\cos A}=2$

(ii) $\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{\cot ^{2} A}{1+\cot ^{2} A}=1$

Sol :

(i) $\frac{\sin ^{3} A+\cos ^{3} A}{\sin A+\cos A}+\frac{\sin ^{3} A-\cos ^{3} A}{\sin A-\cos A}=2$

L.H.S$=\frac{\sin ^{3} A+\cos ^{3} A}{\sin A+\cos A}+\frac{\sin ^{3} A-\cos ^{3} A}{\sin A-\cos A}$

$=\frac{(\sin A+\cos A)\left(\sin ^{2} A-\sin A \cos A+\cos ^{2} A\right)}{(\sin A+\cos A)}+\frac{(\sin A-\cos A)\left(\sin ^{2} A+\sin A \cos A+\cos ^{2} A\right)}{(\sin A-\cos A)}$

$=(1-\sin A \cos A)+(1+\sin A \cos A)$  $\left[\because \sin ^{2} A+\cos ^{2} A=1\right\}$

$=1-\sin A \cos A+1+\sin A \cos A=2$=R.H.S

(ii) $\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{\cot ^{2} A}{1+\cot ^{2} A}=1$

$=\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{\frac{1}{\tan ^{2} A}}{1+\frac{1}{\tan ^{2} A}}$

$=\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{\frac{1}{\tan ^{2} A}}{\frac{\tan ^{2} A+1}{\tan ^{2} A}}$

$=\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{\tan ^{2} A}{\tan ^{2} A\left(\tan ^{2} A+1\right)}$

$=\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{1}{1+\tan ^{2} A}$

$=\frac{1+\tan ^{2} A}{1+\tan ^{2} A}=1$

=R.H.S

Question 30

(i) $\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}$

(ii) $(\sin A+\sec A)^{2}+(\cos A+\operatorname{cosec} A)^{2}=(1+\sec A \quad \operatorname{cosec} A)^{2}$

(iii) $\frac{\tan A+\sin A}{\tan A-\sin A}=\frac{\sec A+1}{\sec A-1}$

Sol :

(i) $\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}$

L.H.S$=\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}$

$=\frac{1}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}}-\frac{1}{\cos A}$

$=\frac{\cos A}{1+\sin A}-\frac{1}{\cos A}$

$=\frac{\cos ^{2} A-1-\sin A}{\cos A(1+\sin A)}=\frac{-\sin ^{2}-\sin A}{\cos A(1+\sin A)}$

$=\frac{-\sin A(1+\sin A)}{\cos A(1+\sin A)}=-\tan A$

R.H.S$=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}$

$=\frac{1}{\cos A}-\frac{1}{\frac{1}{\cos A}-\frac{\sin A}{\cos A}}$

$=\frac{1}{\cos A}-\frac{\cos A}{1-\sin A}$

$=\frac{1-\sin A-\cos ^{2} A}{\cos A(1-\sin A)}$

$=\frac{\sin ^{2} A-\sin A}{\cos A(1-\sin A)}$

$=\frac{-\sin A+\sin ^{2} A}{\cos A(1-\sin A)}$

$=\frac{-\sin A(1-\sin A)}{\cos A(1-\sin A)}$

$\frac{-\sin A}{\cos A}=-\tan A$

∴L.H.S=R.H.S

(ii) $(\sin A+\sec A)^{2}+(\cos A+\operatorname{cosec} A)^{2}=(1+\sec A \operatorname{cosec} A)^{2}$

L.H.S$=(\sin A+\sec A)^{2}+(\cos A+\operatorname{cosec} A)^{2}$

$=\sin ^{2} A+\sec ^{2} A+2 \sin A \sec A+\cos ^{2} A+\operatorname{cosec}^{2} A+2 \cos A \operatorname{cosec} A$

$=\left(\sin ^{2} A+\cos ^{2} A\right)+\left(\sec ^{2} A+\operatorname{cosec}^{2} A\right)+2 \sin A \times \frac{1}{\cos A}+2 \times \cos A \times \frac{1}{\sin A}$

$=1+\left[\frac{1}{\cos ^{2} A}+\frac{1}{\sin ^{2} A}\right]+\frac{2 \sin ^{2} A+2 \cos ^{2} A}{\sin A \cos A}$

$=1+\left[\frac{\sin ^{2} A+\cos ^{2} A}{\cos ^{2} A \sin ^{2} A}\right]+\frac{2\left[\sin ^{2} A+\cos ^{2} A\right]}{\sin A \cos A}$

$=1+\frac{1}{\cos ^{2} A \sin ^{2} A}+\frac{2}{\sin A \cos A}$

$\left[\because \sin ^{2} \theta+\cos ^{2} \theta+1\right]$

$=\left(1+\frac{1}{\cos A \sin A}\right)^{2}$ $\left.\left[\because(a+b)^{2}=a^{2}+(b)^{2}+2 a b\right)\right]$

$=(1+\operatorname{cosec} A \sec A)^{2}$

=R.H.S

(iii) $\frac{\tan A+\sin A}{\tan A-\sin A}=\frac{\sec A+1}{\sec A-1}$

L.H.S$=\frac{\tan A+\sin A}{\tan A-\sin A}$

$=\frac{\frac{\sin A}{\cos A}+\sin A}{\frac{\sin A}{\cos A}-\sin A}$

$=\frac{\frac{\sin A+\sin A \cos A}{\cos A}}{\frac{\sin A-\sin A \cos A}{\cos A}}$

$=\frac{\sin A(1+\cos A)}{\sin A(1-\cos A)}=\frac{1+\cos A}{1-\cos A}$

Dividing each term by cos A

$\frac{\frac{1}{\cos A}+1}{\frac{1}{\cos A}-1}=\frac{\sec A+1}{\sec A-1}$=R.H.S

Question 31

If sin θ + cos θ = √2 sin (90° – θ), show that cot θ = √2 + 1

Sol :

sin θ + cos θ = √2 sin (90° – θ)

sin θ + cos θ = √2 cos θ

dividing by sin θ

$1+\cot \theta=\sqrt{2} \cot \theta$

$1=\sqrt{2} \cot \theta-\cot \theta$

$1=(\sqrt{2}-1) \cot \theta$

$\cot \theta=\frac{1}{\sqrt{2}-1}=\frac{1 \times(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}$ (rationalising the denominator)

$=\frac{(\sqrt{2}+1)}{(\sqrt{2})^{2}-(1)^{2}}=\frac{\sqrt{2}+1}{2-1}=\frac{\sqrt{2}+1}{1}$

$=\sqrt{2}+1$

=R.H.S

Question 32

If $7 \sin ^{2} \theta+3 \cos ^{2} \theta=4,0^{\circ} \leq \theta \leq 90^{\circ}$, then find the value of θ.

Sol :

$7 \sin ^{2} \theta+3 \cos ^{2} \theta=4,0^{\circ} \leq \theta \leq 90^{\circ}$

$3 \sin ^{2} \theta+3 \cos ^{2} \theta+4 \sin ^{2} \theta=4$

$3\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=4-4 \sin ^{2} \theta$

$3 \times 1=4\left(1-\sin ^{2} \theta\right) \Rightarrow \frac{3}{4}=\cos ^{2} \theta$

$\cos \theta=\frac{\sqrt{3}}{2}=\cos 30^{\circ}$

$\therefore \theta=30^{\circ}$

Question 33

If sec θ + tan θ = m and sec θ – tan θ = n, prove that mn = 1.

Sol :

sec θ + tan θ = m and sec θ – tan θ = n

mn = (sec θ + tan θ) (sec θ – tan θ) 

=$\sec ^{2} \theta-\tan ^{2} \theta=1$ 

$\left(\therefore \sec ^{2} \theta-\tan ^{2} \theta=1\right)$

Hence proved.

Question 34

If x – a sec θ + b tan θ and y = a tan θ + b sec θ, prove that $x^{2}-y^{2}=a^{2}-b^{2}$

Sol :

x – a sec θ + b tan θ and y = a tan θ + b sec θ

To prove that $x^{2}-y^{2}=a^{2}-b^{2}$

$=\left(a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta+2 a b \sec \theta \tan \theta\right)-\left(a^{2} \tan ^{2} \theta+b^{2} \sec ^{2} \theta+2 a b \sec \theta \tan \theta\right)$

$=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta+2 a b \sec \theta \tan \theta-a^{2} \tan ^{2} \theta-b^{2} \sec ^{2} \theta-2 a b \sec \theta \tan \theta$

$=a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)-b^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)$

$=a^{2} \times 1-b^{2} \times 1$ $\left\{\sec ^{2} \theta-\tan ^{2} \theta=1\right\}$

$=a^{2}-b^{2}$

Hence proved

Question 35

If x = h + a cos θ and y = k + a sin θ, prove that $(x-h)^{2}+(y-k)^{2}=a^{2}$

Sol :

x = h + a cos θ and y = k + a sin θ

To prove that $(x-h)^{2}+(y-k)^{2}=a^{2}$

$(x-h)=a \cos \theta$

$(y-k)=a \sin \theta$

Squaring and adding,

$(x-h)^{2}+(y-k)^{2}=a^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta$

$=a^{2} \cdot\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$ $\left\{\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

$=a^{2} \times 1=a^{2}$