Without using trigonometric tables, evaluate:

Question:

Without using trigonometric tables, evaluate:

(i) $\frac{\sin 26^{\circ}}{\cos 64^{\circ}}$

(ii) $\frac{\sec 11^{\circ}}{\operatorname{cosec} 79^{\circ}}$

(iii) $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$

(iv) $\frac{\cos 37^{\circ}}{\sin 53^{\circ}}$

(v) $\frac{\operatorname{cosec} 42^{\circ}}{\sec 48^{\circ}}$

(vi) $\frac{\cot 34^{\circ}}{\tan 56^{\circ}}$

Solution:

(i) $\frac{\sin 26^{\circ}}{\cos 64^{\circ}}$

$\frac{\sin 26^{\circ}}{\cos 64^{\circ}}=\frac{\sin \left(90^{\circ}-64^{\circ}\right)}{\cos 64^{\circ}}$

$=\frac{\cos 64^{\circ}}{\cos 64^{\circ}} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right)$

$=1$

Hence, $\frac{\sin 26^{\circ}}{\cos 64^{\circ}}=1$

(ii) $\frac{\sec 11^{\circ}}{\operatorname{cosec} 79^{\circ}}$

$=\frac{\sec \left(90^{\circ}-79^{\circ}\right)}{\operatorname{cosec} 79^{\circ}}$

$=\frac{\operatorname{cosec} 79^{\circ}}{\operatorname{cosec} 79^{\circ}}[\because \sec (90-\theta)=\operatorname{cosec} \theta]$

$=1$

(iii) $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$

$\frac{\tan 65^{\circ}}{\cot 25^{\circ}}=\frac{\tan \left(90^{\circ}-25^{\circ}\right)}{\cot 25^{\circ}}$

$=\frac{\cot 25^{\circ}}{\cot 25^{\circ}} \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right)$

$=1$

Hence, $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}=1$

(iv) $\frac{\cos 37^{\circ}}{\sin 53^{\circ}}$

$\frac{\cos 37^{\circ}}{\sin 53^{\circ}}=\frac{\cos \left(90^{\circ}-53^{\circ}\right)}{\sin 53^{\circ}}$

$=\frac{\sin 53^{\circ}}{\sin 53^{\circ}} \quad\left(\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$

$=1$

Hence, $\frac{\cos 37^{\circ}}{\sin 53^{\circ}}=1$

$(v) \frac{\operatorname{cosec} 42^{\circ}}{\sec 48^{\circ}}$

$=\frac{\operatorname{cosec}\left(90^{\circ}-48^{\circ}\right)}{\sec 48^{\circ}}$

$=\frac{\sec 48^{\circ}}{\sec 48^{\circ}} \quad[\because \sec (90-\theta)=\operatorname{cosec} \theta]$

$=1$

(vi) $\frac{\cot 34^{\circ}}{\tan 56^{\circ}}$

$\frac{\cot 34^{\circ}}{\tan 56^{\circ}}=\frac{\cot \left(90^{\circ}-56^{\circ}\right)}{\tan 56^{\circ}}$

$=\frac{\tan 56^{\circ}}{\tan 56^{\circ}} \quad\left(\because \cot \left(90^{\circ}-\theta\right)=\tan \theta\right)$

$=1$

Hence, $\frac{\cot 34^{\circ}}{\tan 56^{\circ}}=1$