Question

Find the value of $\frac{\sec {15}^o}{co\sec {75}^o}+\frac{\sin {72}^o}{\cos {18}^o}-\frac{\tan {33}^o}{\cot {57}^o}$.

Answer 1

To find:

$\frac{sec{15}^{\circ }}{cosec{75}^{\circ }}+\frac{sin{72}^{\circ }}{cos{18}^{\circ }}-\frac{tan{33}^{\circ }}{cot{57}^{\circ }}=?$

Solution:

$\frac{sec{15}^{\circ }}{cosec{75}^{\circ }}+\frac{sin{72}^{\circ }}{cos{18}^{\circ }}-\frac{tan{33}^{\circ }}{cot{57}^{\circ }}$

$=\frac{sec({90}^{\circ }-{75}^{\circ })}{cosec{75}^{\circ }}+\frac{sin({90}^{\circ }-{18}^{\circ })}{cos{18}^{\circ }}-\frac{tan({90}^{\circ }-{57}^{\circ })}{cot{57}^{\circ }}$

$=\frac{cosec{75}^{\circ }}{cosec{75}^{\circ }}+\frac{cos{18}^{\circ }}{cos{18}^{\circ }}-\frac{cot{57}^{\circ }}{cot{57}^{\circ }}$ $(\because sin({90}^{\circ }-\theta )=cos\theta ,$ $sec({90}^{\circ }-\theta )=cosec\theta ,$ $tan({90}^{\circ }-\theta )=cot\theta ),$

$=1+1-1$

$=2-1$

$=1$