Question

Evaluate : (a) $\frac{\sin {18}^{\circ }}{\cos 72}$ (b) $cosec{31}^{\circ }-\sec {59}^{\circ }$

Answer 1

$a).\frac{\sin 18°}{\cos 72°}$

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

$\Rightarrow \frac{\sin 18}{\cos 72}=\frac{\cos (90-18)}{\cos 72}=\frac{\cos 72}{\cos 72}=1$

$b).\csc 31°-\sec 59°$

$\Rightarrow \csc \theta =\sec (90-\theta )$

$=\sec (90-31°)-\sec 59°$

$=\sec 59°-\sec 59°$

$=0.$

Hence, solved.