SChand CLASS 9 Chapter 19 Trigonometric Ratios Exercise 19(D)

   Exercise 19 D 

Question 1 

Ans:(i) $\frac{\sin 16^{\circ}}{\cos 74^{\circ}}=\frac{\sin 16^{\circ}}{\cos \left(90^{\circ}-16^{\circ}\right)}$

$=\frac{\sin 16^{\circ}}{\sin 16^{\circ}}=1$

(ii) $\frac{\cos 25^{\circ}}{\sin 65^{\circ}}=\frac{\cos 25^{\circ}}{\sin \left(90^{\circ}-25^{\circ}\right)}$

$=\frac{\cos 25^{\circ}}{\cos 25^{\circ}}=1$

(iii)

$\begin{aligned} \frac{\tan 38^{\circ}}{\cot 52^{\circ}} &=\frac{\tan 38^{\circ}}{\cot \left(90^{\circ}-38^{\circ}\right)} \\ &=\frac{\tan 38^{\circ}}{\tan 38^{\circ}}=1 \\ \text { (iv) } \frac{\sec 62^{\circ}}{\operatorname{cosec} 28^{\circ}} &=\frac{\sec 62^{\circ}}{\operatorname{cosec}\left(90^{\circ}-62^{\circ}\right)} \end{aligned}$

$=\frac{\sec 62^{\circ}}{\sec 62^{\circ}}=1$

Question 2 

Ans:(i)

$\begin{aligned} & \sin ^{2} 67^{\circ}+\sin ^{2} 23^{\circ} \\=& \sin ^{2} 67^{\circ}+\sin ^{2}\left(90^{\circ}-67^{\circ}\right) \\=& \sin ^{2} 67^{\circ}+\cos ^{2} 67^{\circ} \end{aligned}$

(ii) $\left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2}$

$=\left(\frac{\sin 49^{\circ}}{\cos \left(90^{\circ}-49^{\circ}\right)}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin \left(90^{\circ}-41^{\circ}\right)}\right)^{2}$

$=\left(\frac{\sin 49^{\circ}}{\sin 49^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\cos 41^{\circ}}\right)^{2}=\left(11^{2}+(1)^{2}\right.$

=1+1=2

(iii) $\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}$

$=\frac{\cos ^{2}\left(90^{\circ}-70^{\circ}\right)+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2}\left(90^{\circ}-590^{2}\right)}$

$=\frac{\sin ^{2} 70^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\cos ^{2} 59^{\circ}}$

$=\frac{1}{1}=1$

(iv) $\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8 \sin ^{2} 30^{\circ}$

$=\frac{\cos \left(90^{\circ}-20^{\circ}\right)}{\sin 20^{\circ}}+\frac{\cos \left(90^{\circ}-31^{\circ}\right)}{\sin 31^{\circ}}-8\left(\sin 30^{\circ}\right)^{2}$

$=\frac{\sin 20^{\circ}}{\sin 20^{\circ}}+\frac{\sin 31^{\circ}}{\sin 31^{\circ}}-8\left(\frac{1}{2}\right)^{2}$

$=1+1-8 \times \frac{1}{4}=2-2=0$

(v) $2 \frac{\tan 33^{\circ}}{\operatorname{coc} 97^{\circ}}-\frac{\cot 80^{\circ}}{\tan 10^{\circ}}$

$=2 \frac{\tan \left(90^{\circ}-37^{\circ}\right)}{\cot 37^{\circ}}-\frac{\cot 80^{\circ}}{\tan \left(90^{\circ}-80^{\circ}\right)}$

$=2 \frac{\cot 37^{\circ}}{\cot 3^{\circ}}-\frac{\cot 80^{\circ}}{\cot 80^{\circ}}$

$=2 \times 1-1=2-1=1$

(vi) $\sec 50^{\circ} \sin 40^{\circ}+\cos 40^{\circ} \operatorname{cosec} 50^{\circ}$

$\begin{aligned}&\frac{\sin 40^{\circ}}{\cos 50^{\circ}}+\frac{\cos 40^{\circ}}{\sin 50^{\circ}} \\&=\frac{\sin 40^{\circ}}{\cos \left(90^{\circ}-40^{\circ}\right)}+\frac{\cos 40^{\circ}}{\sin\left(90^{\circ}-40^{\circ}\right)} \\&=\frac{\sin 40^{\circ}}{\sin 40^{\circ}}+\frac{\cos 40^{\circ}}{\cos 40^{\circ}} \\&=1+1=2\end{aligned}$

(vii) 

$\begin{aligned} & \frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ} \\=& \frac{\cos 80^{\circ}}{\sin \left(10^{\circ}\right)}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}} \\=& \frac{\cos 80^{\circ}}{\sin \left(90^{\circ}-80^{\circ}\right)}+\frac{\cos 59^{\circ}}{\sin \left(90^{\circ}-59^{\circ}\right)} \\=& \frac{\cos 80^{\circ}}{\cos 80^{\circ}}+\frac{\cos 59^{\circ}}{\cos 59^{\circ}} \end{aligned}$

$=1+1=2$

(viii)

 $\begin{aligned} & \frac{2 \sin 43^{\circ}}{\cos 47^{\circ}}-\frac{\cot 30^{\circ}}{\tan 60^{\circ}}-\sqrt{2} \sin 45^{\circ} \\=& 2 \frac{\sin 43^{\circ}}{\cos \left(90^{\circ}-43^{\circ}\right)}-\frac{\cot \left(90^{\circ}-60^{\circ}\right)}{\tan 60^{\circ}}-\sqrt{2} \times \frac{1}{\sqrt{2}} \\=& 2 \frac{\sin 43^{\circ}}{\sin 43^{\circ}}-\frac{\tan 60^{\circ}}{\tan 60^{\circ}}-\sqrt{2} \times \frac{1}{\sqrt{2}} \\=& 2 \times 1-1-1=2-2=0 \end{aligned}$

(ix) $\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 20^{\circ}+\sin ^{2} 70^{\circ}}+\sin ^{2} 64^{\circ}+\cos 64^{\circ} \sin 26^{\circ}$

$=\frac{\cos ^{2}\left(90^{\circ}-70^{\circ}\right)+\cos ^{2} 70^{\circ}}{\sin ^{2} 20^{\circ}+\sin ^{2}\left(90^{\circ}-20^{\circ}\right)}$ +$\sin ^{2} 64+\cos 610 \sin \left(90^{\circ}\right.$ -64^{\circ}$

$=\frac{\sin ^{2} 70^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 20^{\circ}+\cos ^{2} 20^{\circ}}+\sin ^{2} 64^{\circ}+\cos 64^{\circ} \cdot \cos 64^{\circ}$

$=\frac{1}{1}+\left(\sin ^{2} 64^{\circ}+\cos ^{2} 64^{\circ}\right)$

$=1+1=2$

Question 3

Ans: Using sin $\left(90^{\circ}-\theta\right) \operatorname{cosec}\left(90^{\circ}-\theta\right)$ and $\cot \left(90^{\circ}-\theta\right)$

(i) $\operatorname{cosec} 69^{\circ}+\cot 69^{\circ}$

$\begin{aligned}&=\operatorname{cosec}\left(90^{\circ}-21^{\circ}\right)+\cot \left(90^{\circ}-21^{\circ}\right) \\&=\sec 21^{\circ}+\tan 21^{\circ}\end{aligned}$

(ii) 

$\begin{aligned} & \sin 85^{\circ}+\operatorname{cosec} 85^{\circ} \\=& \sin \left(90^{\circ}-5^{\circ}\right)+\operatorname{cosec}\left(90^{\circ}-5^{\circ}\right) \\=& \cos 5^{\circ}+\sec 5^{\circ} \end{aligned}$

Question 4

Ans: 

(i) 

$\begin{aligned} & \frac{\tan \left(90^{\circ}-A\right)}{\operatorname{cosec} A}=\cos A \\ \text { L.H.15. } &=\frac{\tan \left(90^{\circ}-A\right)}{\operatorname{cosec} A}=\frac{\cot A}{\operatorname{cosec} A} \\ &=\frac{\cos A \times \sin A}{\sin A \times 1}=\cos A=\text { R.HS } \end{aligned}$

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

LHS $=\frac{\cos \theta}{\sin \theta\left(90^{\circ}-\theta\right.}+\frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)}$

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

$=1+1=2=$ R.H.S

(iii)$\sec \left(90^{\circ}-\theta\right) \operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec ^{2} \theta \operatorname{coc} \theta$

LH.S $=\sec \left(90^{\circ}-\theta\right) \operatorname{cosec}\left(90^{\circ}-0\right)$

$=\operatorname{cosec} \theta \sec \theta$

$R \cdot H \cdot S=\sec ^{2} \theta \cot \theta=\sec ^{2} \theta \times \frac{\cos \theta}{\sin \theta}$

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

$=\sec \theta \operatorname{cosec} \theta=\operatorname{cosec} \theta \sec \theta$

so L.H.S $=\mathrm{RH} . \mathrm{S}$

(iv)$\sin \left(60^{\circ}-\theta\right)=\cos \left(30^{\circ}+\theta\right)$

LR.S $=\sin \left(60^{\circ}-\theta\right)=\sin \left\{90^{\circ}-\left(30^{\circ}+\theta\right)\right\}$

$=\cos \left(30^{\circ}+\theta\right)$

= RHS

(v) $\frac{\sin 0}{\sin \left(90^{\circ}-\theta\right)}+\frac{\cos \theta}{\cos \left(90^{\circ}-0\right)}=\sec \left(90^{\circ}-0\right)$

$\operatorname{cosec}\left(90^{\circ}-\theta\right)$

$\text { L.H.S }=\frac{\sin \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\cos \theta}{\cos\left(90^{\circ}-0\right)}$

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

$=\operatorname{cosec} \theta \sec \theta$

RHS $=\sec \left(90^{\circ}-\theta\right) \operatorname{cosec}\left(90^{\circ}-\theta\right)$

$=\operatorname{cosec} \theta \sec \theta$

So L.H.S. $=$ R.H.S

(vi)$\cos \left(81^{\circ}+\theta )=\sin \left(9^{\circ}-\theta\right)\right.$

$\begin{aligned} \operatorname{LHS} &=\cos \left(81^{\circ}+\theta\right)=\cos \left\{\left(90^{\circ}-9^{\circ}\right)+\theta\right\} \\ &=\cos \left[90^{\circ}-\left(9^{\circ}-\theta\right)\right\}=\sin 9^{\circ}-\theta \\ &=R . H S \cdot \quad\left\{\cos \left(90^{\circ}-\theta\right)=\sin \theta\right\} \end{aligned}$

Question 5

Ans: (i) $\sin \left(\theta+36^{\circ}\right)=\cos \theta$

$\Rightarrow \sin \left(\theta+36^{\circ}\right)=\sin \left(90^{\circ}-\theta\right)$ $\left\{\sin \left(90^{\circ}-\theta\right)=\cos \theta\right\}$

So $\theta+36^{\circ}=90^{\circ}-\theta \Rightarrow \theta+\theta=90^{\circ}-36^{\circ}$

$=20=54^{\circ} \Rightarrow \theta=\frac{54^{\circ}}{2}=27^{\circ}$

So $\theta=27^{\circ}$

(ii) $\sin 50=\cos 40$

$\Rightarrow \sin 50=\sin \left(90^{\circ}-40\right)$ {cosθ = sin (90 -θ)}

So 5.= 90 -40= 50 +40 =90

$\Rightarrow 30=90^{\circ} \Rightarrow θ=\frac{90^{\circ}}{9}=10^{\circ}$

So θ =$10^{\circ}$

(iii) 

$\begin{aligned} \sin 3 A &=\cos 2 A \\ \sin 3 A &=\sin \left(90^{\circ}-2 A\right) \\\{\cos 0&\left.=\sin \left(90^{\circ}-0\right)\right\} \end{aligned}$

So $3 A=90^{\circ}-2 A \Rightarrow 3 A+2 A=90^{\circ}$

$\Rightarrow 5 A=90^{\circ} \Rightarrow A=\frac{90^{\circ}}{5^{\circ}}=18^{\circ}$

So $A=18^{\circ}$