<s>[INST] How do I go about simplifying $1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1}$ to $\sin x\cos x$ 
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?
 [/INST] We have
$$1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1} =1-\frac{\sin^2x\tan x+\cos^2 x}{\tan x+1}=1-\frac{\sin^3x+\cos^3 x}{\sin x + \cos x}=\\=\frac{\sin x(1-\sin^2x)+\cos x(1-\cos^2 x)}{\sin x + \cos x}=\frac{\sin x\cos^2x+\cos x\sin^2 x}{\sin x + \cos x}=\sin x \cos x$$
</s>