Question

What will be the value of $(\frac{\sin {50}^0}{\cos {40}^0}+\frac{cosec{40}^0}{\sec {50}^0}-4\cos {50}^{0}cosec{40}^0)?$

Answer 1

We have,

$\frac{\sin {50}^0}{\cos {40}^0}+\frac{\csc {40}^0}{\sec {50}^0}-4\cos {50}^0\csc {40}^0$

$=\frac{\sin ({90}^0-{40}^0)}{\cos {40}^0}+\frac{\csc ({90}^0-{50}^0)}{\sec {50}^0}-4\cos {50}^0\csc ({90}^0-{50}^0)$

$=\frac{\cos {40}^0}{\cos {40}^0}+\frac{\sec {50}^0}{\sec {50}^0}-4\cos {50}^0\sec {50}^0$

$=1+1-4\cos {50}^0\cdot \frac{1}{\cos {50}^0}$

$=2-4=-2$

Hence, this is the answer.