Question

Evaluate : $\frac{\sec x^{\circ }+\tan x^{\circ }}{\sec x^{\circ }-\tan x^{\circ }}$

Answer 1

$\begin{array}{c}\frac{{secx}^{\circ }+\tan x^{\circ }}{\sec x^{\circ }-\tan x^{\circ }}\\ =\frac{1+\sin x^{\circ }}{1-{sinx}^{\circ }}\\ =\frac{{\left(\cos \frac{x^{\circ }}{2}+\sin \frac{x^{\circ }}{2}\right)}^2}{{\left(\cos \frac{x^{\circ }}{2}-\sin \frac{x^{\circ }}{2}\right)}^2}\\ ={\left(\frac{1+\tan \frac{x^{\circ }}{2}}{1-\tan \frac{x^{\circ }}{2}}\right)}^2\\ ={\left[\tan \left(\frac{\pi }{4}+\frac{x^{\circ }}{2}\right)\right]}^2\\ ={\tan }^2\left(\frac{\pi }{2}+\frac{x^{\circ }}{2}\right)\end{array}$