Question

Solve:$\frac{\tan x+\cot x}{\tan x-\cot x}$

Answer 1

Given,

$\frac{\tan x+\cot x}{\tan x-\cot x}=\frac{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}{\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}}=\frac{\frac{{\sin }^{2}x+{\cos }^{2}x}{\cos x\sin x}}{\frac{{\sin }^{2}x-{\cos }^{2}x}{\cos x\sin x}}=\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x-{\cos }^{2}x}=\frac{1}{2{\sin }^{2}x-1}[Since,{\sin }^{2}x+{\cos }^{2}x=1,{\sin }^{2}x-{\cos }^{2}x=2{\sin }^{2}x-1]\therefore \frac{\tan x+\cot x}{\tan x-\cot x}=\frac{1}{2{\sin }^{2}x-1}.$