# How do you verify (1+tan^2x)/(1+cot^2x) = (tan^2x)^2?

Jul 20, 2016

These are called the Trigonometric Pythagorean Identities

#### Explanation:

Suppose you have a right triangle, with sides $a$ and $b$, and hypotenuse $c$. Now, suppose that $a$ is the side adjacent to the working angle. Then, by Pythagorean Theorem, you have ${a}^{2} + {b}^{2} = {c}^{2}$.

$1 + {\tan}^{2} x = {\sec}^{2} x$
$1 + {\cot}^{2} x = {\csc}^{2} x$

The first identity comes from dividing the equation by ${a}^{2}$, and the second by ${b}^{2}$. And remembering the basic trigonometric identities: $S o h , C a h , T o a$. There is a third identity, but I will leave that as an exercise for you, if you are interested.

Now, with those identities, we can substitute in the original expression:
$\frac{1 + {\tan}^{2} x}{1 + {\cot}^{2} x} = \frac{{\sec}^{2} x}{{\csc}^{2} x} = \frac{{\sin}^{2} x}{{\cos}^{2} x} = {\left({\tan}^{2} x\right)}^{2}$

I believe the last three parts of the equalities are well known facts, or at least extremely easy to show, hence I'll stop here with the proof.