Question

Evaluate :
$\frac{\sin {50}^{\circ }}{\cos {40}^{\circ }}+\frac{cosec{40}^{\circ }}{\sec {50}^{\circ }}-4\cos {50}^{\circ }cosec{40}^{\circ }$

Answer 1

We know that,

$\cos (90-\theta )=\sin \theta$

$\sec (90-\theta )=cosec\theta$

$cosec\theta =\frac{1}{\sin \theta }$

Therefore,

$\cos {40}^{\circ }=\cos ({90}^{\circ }-{50}^{\circ })=\sin {50}^{\circ }$

$\sec {50}^{\circ }=\tan ({90}^{\circ }-{40}^{\circ })=cosec{40}^{\circ }$

$\cos {50}^{\circ }=\sin ({90}^{\circ }-{40}^{\circ })=\sin {40}^{\circ }$

To evaluate,

$\frac{\sin {50}^{\circ }}{\cos {40}^{\circ }}+\frac{\csc {40}^{\circ }}{\sec {50}^{\circ }}-4\cos {50}^{\circ }cosec{40}^{\circ }$

$=\frac{\sin {50}^{\circ }}{\sin {50}^{\circ }}+\frac{cosec{40}^{\circ }}{cosec{40}^{\circ }}-4\left(\frac{\cos {50}^{\circ }}{\sin {40}^{\circ }}\right)$

$=1+1-4\left(\frac{\sin {40}^{\circ }}{\sin {40}^{\circ }}\right)$

$=1+1-4$

$=-2$