Question

((1-cos^2)/tan^2)+2sin^2

Answer 1

`(1-\cos^2x)/(\tan^2x)+2\sin^2x\\\\\text{use}\ \tan x=(\sin x)/(\cos x)\ \text{and}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\=(\sin^2x)/(\left((\sin x)/(\cos x)\right)^2)+2\sin^2x=(\sin^2x)/((\sin^2x)/(\cos^2x))+2\sin^2x=\sin^2x\cdot(\cos^2x)/(\sin^2x)+2\sin^2x\\\\=\cos^2x+2\sin^2x=\cos^2x+2(1-\cos^2x)\\\\\text{use distributive property}\\\\=\cos^2x+2-2\cos^2x=2-\cos^2x`