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