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