Question

Verify the identity tan x + cot x / tan x - cot x = 1/ sin^2x - cos^2x

Answer 1

$\dfrac{\tan x+\cot x}{\tan x-\cot x}=\dfrac{1}{\sin^2x-\cos^2x}\\\\\text{use}\ \tan x=\dfrac{\sin x}{\cos x},\ \cot x=\dfrac{\cos x}{\sin x}\\\\\tan x+\cot x=\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}=\dfrac{\sin x\cdot\sin x}{\sin x\cos x}+\dfrac{\cos x\cdot\cos x}{\sin x\cos x}\\\\=\dfrac{\sin^2x+\cos^2x}{\sin x\cos x}\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=\dfrac{1}{\sin x\cos x}\\\\\tan x-\cot x=\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\sin x}=\dfrac{\sin^2x-\cos^2x}{\sin x\cos x}$

$L_s=\dfrac{\tan x+\cot x}{\tan x-\cot x}=\dfrac{\frac{1}{\sin x\cos x}}{\frac{\sin^2x-\cos^2x}{\sin x\cos x}}=\dfrac{1}{\sin x\cos x}\cdot\dfrac{\sin x\cos x}{\sin^2x-\cos^2x}\\\\=\dfrac{1}{\sin^2x-\cos^2x}=R_s\\\\L_s=R_s\Rightarrow The\ identity.$