Answer
$$\sin^2x(1+\cot x)+\cos^2x(1-\tan x)+\cot^2x=\csc^2x$$
We prove the expression to be an identity by simplifying the left side.
Work Step by Step
$$\sin^2x(1+\cot x)+\cos^2x(1-\tan x)+\cot^2x=\csc^2x$$
We start with the left side.
$$A=\sin^2x(1+\cot x)+\cos^2x(1-\tan x)+\cot^2x$$
1) First, $\cot x=\frac{\cos x}{\sin x}$
So, $$\sin^2x(1+\cot x)=\sin^2x(1+\frac{\cos x}{\sin x})=\sin^2x\times\frac{\sin x+\cos x}{\sin x}=\sin x(\sin x+\cos x)$$
2) Second, $\tan x=\frac{\sin x}{\cos x}$
So, $$\cos^2x(1-\tan x)=\cos^2x(1-\frac{\sin x}{\cos x})=\cos^2x\times\frac{\cos x-\sin x}{\cos x}=\cos x(\cos x-\sin x)$$
Now applying them to $A$:
$$A=\sin x(\sin x+\cos x)+\cos x(\cos x-\sin x)+\cot^2x$$
$$A=\sin^2x+\sin x\cos x+\cos^2 x-\sin x\cos x+\cot^2x$$
$$A=\sin^2 x+\cos^2x+\cot^2x $$
$$A=1+\cot^2x$$ (Pythagorean Identity: $\sin^2x+\cos^2x=1$)
$$A=\csc^2x$$ (Pythagorean Identity: $1+\cot^2x=\csc^2x$)
Therefore, 2 sides are equal and the expression is an identity.
