Question

Verify the identity (tan x + 1)^2 + (tan x-1)^2= 2 sec^2 x

Answer 1

`(\tan x+1)^2+(\tan x-1)^2=2\sec^2x\\\\\text{use}\ \tan x=(\sin x)/(\cos x)\\\\L_s=\left((\sin x)/(\cos x)+(\cos x)/(\cos x)\right)^2+\left((\sin x)/(\cos x)-(\cos x)/(\cos x)\right)^2\\\\=\left((\sin x+\cos x)/(\cos x)\right)^2+\left((\sin x-\cos x)/(\cos x)\right)^2\\\\=((\sin x+\cos x)^2)/(\cos^2x)+((\sin x-\cos x)^2)/(\cos^2x)\\\\\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2`

`=(\sin^2x+2\sin x\cos x+\cos^2)/(\cos^2x)+(\sin^2x-2\sin x\cos x+\cos^2)/(\cos^2x)\\\\=(\sin^2x+2\sin x\cos x+\cos^2+\sin^2x-2\sin x\cos x+\cos^2)/(\cos^2x)\\\\=(2\sin^2x+2\cos^2x)/(\cos^2x)=(2(\sin^2x+\cos^2x))/(\cos^2x)\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=(2(1))/(\cos^2x)=2\cdot(1)/(\cos^2x)=2\left((1)/(\cos x)\right)^2\\\\\text{use}\ \sec x=(1)/(\cos x)\\\\=2(\sec^2x)=2\sec^2x=R_s\\\\L_s=R_s\Rightarrow The\ identity`