# Evaluate the following :

Question:

Evaluate the following :

(i) $\frac{\sin 20^{\circ}}{\cos 70^{\circ}}$

(ii) $\frac{\cos 19^{\circ}}{\sin 71^{\circ}}$

(iii) $\frac{\sin 21^{\circ}}{\cos 69^{\circ}}$

(iv) $\frac{\tan 10^{\circ}}{\cot 80^{\circ}}$

(V) $\frac{\sec 11^{\circ}}{\operatorname{cosec} 79^{\circ}}$

Solution:

(i) Given that $\frac{\sin 20}{\cos 70}$

Since $\sin (90-\theta)=\cos \theta$

$\Rightarrow \frac{\sin 20}{\cos 70}=\frac{\sin (90-70)}{\cos 70}$

$\Rightarrow \frac{\sin 20}{\cos 70}=\frac{\cos 70}{\cos 70}$

$\Rightarrow \frac{\sin 20}{\cos 70}=1$

Therefore $\frac{\sin 20}{\cos 70}=1$

(ii) Given that $\frac{\cos 19}{\sin 71}$

$\Rightarrow \frac{\cos 19}{\sin 71}=\frac{\cos (90-71)}{\sin 71}$

$\Rightarrow \frac{\cos 19}{\sin 71}=\frac{\sin 71}{\sin 71}$

$\Rightarrow \frac{\cos 19}{\sin 71}=1$

Since $\cos (90-\theta)=\sin \theta$

Therefore $\frac{\cos 19}{\sin 71}=1$

(iii) Given that $\frac{\sin 21}{\cos 69}$

Since $\sin (90-\theta)=\cos \theta$

$\Rightarrow \frac{\sin 21}{\cos 69}=\frac{\sin (90-69)}{\cos 69}$

$\Rightarrow \frac{\sin 21}{\cos 69}=\frac{\cos 69}{\cos 69}$

$\Rightarrow \frac{\sin 21}{\cos 69}=1$

(iv) We are given that $\frac{\tan 10}{\cot 80}$

Since $\tan (90-\theta)=\cot \theta$

$\Rightarrow \frac{\tan 10}{\cot 80}=\frac{\tan (90-80)}{\cot 80}$

$\Rightarrow \frac{\tan 10}{\cot 80}=\frac{\cot 80}{\cot 80}$

$\Rightarrow \frac{\tan 10}{\cot 80}=1$

Therefore $\frac{\tan 10}{\cot 80}=1$

(v) Given that $\frac{\sec 11}{\operatorname{cosec} 79}$

Since $\sec (90-\theta)=\operatorname{cosec} \theta$

$\Rightarrow \frac{\sec 11}{\operatorname{cosec} 79}=\frac{\sec (90-79)}{\operatorname{cosec} 79}$

$\Rightarrow \frac{\sec 11}{\operatorname{cosec} 79}=\frac{\operatorname{cosec} 79}{\operatorname{cosec} 79}$

$\Rightarrow \frac{\sec 11}{\operatorname{cosec} 79}=1$

Therefore $\frac{\sec 11}{\operatorname{cosec} 79}=1$