Without using trigonometric tables, evaluate the following:

Question:

Without using trigonometric tables, evaluate the following:

$\frac{\sec 37^{\circ}}{\operatorname{cosec} 53^{\circ}}+2 \cot 15^{\circ} \cot 25^{\circ} \cot 45^{\circ} \cot 75^{\circ} \cot 65^{\circ}-3\left(\sin ^{2} 18^{\circ}+\sin ^{2} 72^{\circ}\right)$

Solution:

We have to evaluate the following expression

$\frac{\sec 37^{\circ}}{\operatorname{cosec} 53^{\circ}}+2 \cot 15^{\circ} \cot 25^{\circ} \cot 45^{\circ} \cot 75^{\circ} \cot 65^{\circ}-3\left(\sin ^{2} 18^{\circ}+\sin ^{2} 72^{\circ}\right)$

Here we are going to use following identities.

$\sec \theta=\operatorname{cosec}(90-\theta), \cot \theta=\tan (90-\theta), \sin \theta=\cos (90-\theta)$ and $\sin ^{2} \theta+\cos ^{2} \theta=1$

The given expression can be written as

$\frac{\sec 37^{\circ}}{\operatorname{cosec} 53^{\circ}}+2\left(\cot 15^{\circ} \cot 75^{\circ}\right)\left(\cot 25^{\circ} \cot 65^{\circ}\right) \cot 45^{\circ}-3\left(\sin ^{2} 18^{\circ}+\sin ^{2} 72^{\circ}\right)$

$=\frac{\operatorname{cosec}\left(90^{\circ}-53^{\circ}\right)}{\operatorname{cosec} 53^{\circ}}+2\left\{\cot 15^{\circ} \tan \left(90^{\circ}-15^{\circ}\right)\right\}\left\{\cot 25^{\circ} \tan \left(90^{\circ}-25^{\circ}\right)\right\} \cot 45^{\circ}$

$-3\left\{\sin ^{2} 18^{\circ}+\cos ^{2}\left(90^{\circ}-18^{\circ}\right)\right\}$

$=\frac{\operatorname{cosec} 53^{\circ}}{\operatorname{cosec} 53^{\circ}}+2\left(\cot 15^{\circ} \tan 15^{\circ}\right)\left(\cot 25^{\circ} \tan 25^{\circ}\right) \cot 45^{\circ}-3\left(\sin ^{2} 18^{\circ}+\cos ^{2} 18^{\circ}\right)$

$=1+2(1)(1)(1)-3(1)$

$=0$

Hence the value of the given expression is zero.