Evaluate :

Solution:

(i) $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}=\frac{\sin 18^{\circ}}{\cos \left(90^{\circ}-18^{\circ}\right)}=\frac{\sin 18^{\circ}}{\sin 18^{\circ}}=1$.

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

(ii) $\frac{\tan 26^{\circ}}{\cot 64^{\circ}}=\frac{\tan \left(80^{\circ}-64^{\circ}\right)}{\cot 64^{\circ}}=\frac{\cot 64^{\circ}}{\cot 64^{\circ}}=1$

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

$=\sin 42^{\circ}-\sin 42^{\circ}=0$

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

= cosec (90° – 59°) – sec 59°

= sec 59° – sec 59° = 0