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) $\frac{\sec 11^{\circ}}{\operatorname{cosec} 79^{\circ}}$
(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 $\frac{\sec 11}{\operatorname{cosec} 79}=1$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.