Question

Solve : $\frac{\sec x^o+\tan x^o}{\sec x^o-\tan x^o}$

Answer 1

$\frac{\sec x°+\tan x°}{\sec x°-\tan x°}=\frac{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}{\frac{1}{\cos x}-\frac{\sin x}{\cos x}}$

$=\frac{1+\sin x}{1-\sin x}\times \frac{(1+\sin x)}{(1+\sin x)}$

$=\frac{{(1+\sin x)}^2}{1-{\sin }^{2}x}$

$=\frac{1+{\sin }^{2}x+2\sin x}{{\cos }^{2}x}$

$=\frac{1}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}+\frac{2\sin x}{{\cos }^{2}x}$

$={\sec }^{2}x+{\tan }^{2}x+2\tan x.\sec x$

$={(\sec x+\tan x)}^2$

Hence, the answer is ${(\sec x+\tan x)}^2$