Evaluate:
(i) $ \frac{\sin 18^{\circ}}{\cos 72^{\circ}} $
(ii) $ \frac{\tan 26^{\circ}}{\cot 64^{\circ}} $
(iii) $ \cos 48^{\circ}-\sin 42^{\circ} $
(iv) $ \operatorname{cosec} 31^{\circ}-\sec 59^{\circ} $.

AcademicMathematicsNCERTClass 10

To do:

We have to evaluate the given expressions.

Solution:  

(i) We know that,

$cos\ (90^{\circ}- \theta) = sin\ \theta$

Therefore,

$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}=\frac{\sin 18^{\circ}}{\cos( 90^{\circ}-18^{\circ})}$

$=\frac{\sin 18^{\circ}}{\sin 18^{\circ}}$

$=1$

Hence, $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}=1$.   

(ii) We know that,

$cot\ (90^{\circ}- \theta) = tan\ \theta$

Therefore,

$\frac{\tan 26^{\circ}}{\cot 64^{\circ}}=\frac{\tan 26^{\circ}}{\cot (90^{\circ}-26^{\circ})}$

$=\frac{\tan 26^{\circ}}{\tan 26^{\circ}}$

$=1$

Therefore, $\frac{\tan 26^{\circ}}{\cot 64^{\circ}}=1$.  

(iii) We know that,

$sin\ (90^{\circ}- \theta) = cos\ \theta$

Therefore,

$\cos 48^{\circ}-\sin 42^{\circ}=\cos 48^{\circ}-\sin (90^{\circ}-48^{\circ})$

$=\cos 48^{\circ}-\cos 48^{\circ}$

$=0$

Therefore, $\cos 48^{\circ}-\sin 42^{\circ}=0$.   

(iv) We know that,

$sec (90^{\circ}- \theta) = cosec\ \theta$

Therefore,

$\operatorname{cosec} 31^{\circ}-\sec 59^{\circ}=\operatorname{cosec} 31^{\circ}-\sec (90^{\circ}-31^{\circ})$

$=\operatorname{cosec} 31^{\circ}-\operatorname{cosec} 31^{\circ}$

$=0$

Therefore, $\operatorname{cosec} 31^{\circ}-\sec 59^{\circ}=0$.   

raja
Updated on 10-Oct-2022 13:22:36

Advertisements