Verifying identity: sin^2x tanx = tanx - cosx sinx

Answer 1

I chose to manipulate the left side of the equation.

Start out substituting in (1-cos²x) in for sin²x using the Pythagorean identity

(1-cos²x)tan x Distribute tan x

tan x - tan x(cos²x) Substitute in tan x = sin x/cos x

tan x - (sin x/cos x)(cos²x) Cancel out one of the cos x

tan x = cos x sin x

Answer 2

convert the tanx to sinx/cosx

sin^2x(sinx/cosx) = sinx/cosx - cos^2xsinx/cosx = (sinx-cos^2xsinx)/cosx = sinx/cosx[1-cos^2x] = sin^3x/cosx

sin3x/cosx = sin^3x/cosx

If you can't see what to do, generally it works to convert everything into sines and cosines and usually use sin^2x + cos^2x = 1 at some point. that was the bracket term 1-cos^2x = sin^2x giving sin^3x/cosx