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,
$(x-h)^{2}+(y-k)^{2}=a^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta$
$=a^{2} \cdot\left(\sin ^{2} \theta+\cos ^{2}
\theta\right)$ $\left\{\because \sin ^{2} \theta+\cos ^{2}
\theta=1\right)$
$=a^{2} \times 1=a^{2}$
