Q:

Simplify these expressions

$$\frac{1}{(1+\tan^2x)}+\frac{1}{(1+\cot^2x)}$$
It resultet to me that it is $\cos^2x+\sin^2x=1$ Am I right?
The other is:
$$\frac{\tan x}{(1-\tan^2x)}\times \frac{(\cot^2x-1)}{\cot x}$$. It result to me that t is $\sin^2x$, but it is not correct

A:

Yes, more generally, for $a\ne-1$,
$$
\frac{1}{1+a}+\frac{1}{1+a^{-1}}=\frac{1}{1+a}+\frac{1}{1+\dfrac{1}{a}}
=\frac{1}{1+a}+\frac{a}{a+1}=1
$$
For the second one, consider, for $a\ne-1$, $a\ne0$ and $a\ne1$,
$$
\frac{a}{1-a^2}\frac{a^{-2}-1}{a^{-1}}=
\frac{a}{1-a^2}\frac{\dfrac{1}{a^2}-1}{\dfrac{1}{a}}=
\frac{a}{1-a^2}\frac{1-a^2}{a^2}\,a=1
$$

