Question

Without using trigonometric tables, evaluate the following:
$2\left(\frac{\cos {58}^{\circ }}{\sin {32}^{\circ }}\right)-\sqrt{3}\left(\frac{\cos {38}^{\circ }\text{cosec}{52}^{\circ }}{\tan {15}^{\circ }\tan {60}^{\circ }\tan {75}^{\circ }}\right)$.

Answer 1

• Before solving this , we need to know that

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

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

$\tan ({90}^{\circ }-\theta )=\cot \theta$

$\tan \theta =\frac{1}{\cot \theta }$

• $2\left(\frac{\cos {58}^{\circ }}{\sin {32}^{\circ }}\right)-\sqrt{3}\left(\frac{\cos {38}^{\circ }\text{cosec}{52}^{\circ }}{\tan {15}^{\circ }\tan {60}^{\circ }\tan {75}^{\circ }}\right)$

or using the above identities

$2\left(\frac{\sin {32}^{\circ }}{\sin {32}^{\circ }}\right)-\sqrt{3}\left(\frac{\sin {52}^{\circ }}{\sin {52}^{\circ }\tan {15}^{\circ }\times \sqrt{3}\times \cot {15}^{\circ }}\right)$

$\text{or}2-1=1$ (every term in brackets cancels out according to the mentioned identities)

• So, $2\left(\frac{\cos {58}^{\circ }}{\sin {32}^{\circ }}\right)-\sqrt{3}\left(\frac{\cos {38}^{\circ }\text{cosec}{52}^{\circ }}{\tan {15}^{\circ }\tan {60}^{\circ }\tan {75}^{\circ }}\right)=1$