How do I go about simplifying $1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1}$ to $\sin x\cos x$

Question

This is a tricky trig problem I'm stuck with. The problem is asking me to simplify $$1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1}$$ to $\sin x\cos x$.

What I've been doing so far is trying to remove those $tan$ functions.
$$1-\left(\frac{\sin^2x\tan x-\cos^2}{\tan x+1}\right)$$ $$1-\left(\frac{\frac{\sin^3x-\cos^3x}{\cos x}}{\tan x+1}\right)$$ $$1-\left(\frac{\frac{\sin^3x-\cos^3x}{\cos x}}{\frac{\sin x}{\cos x}+1}\right)$$ I made this complicated. Is there a simple way to do this problem?

Answer 1

We have

$$1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1} =1-\frac{\sin^2x\tan x+\cos^2 x}{\tan x+1}=1-\frac{\sin^3x+\cos^3 x}{\sin x + \cos x}=\\=\frac{\sin x(1-\sin^2x)+\cos x(1-\cos^2 x)}{\sin x + \cos x}=\frac{\sin x\cos^2x+\cos x\sin^2 x}{\sin x + \cos x}=\sin x \cos x$$

Answer 2

You dropped a sign in the first line. The $\cos^2 x$ term should be positive inside the parentheses. Now multiply the numerator and denominator by $\cos x$ and you have a sum of cubes in the numerator. The denominator cancels.