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

1 Answer
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^2x=sec^2x#
#1+cot^2x=csc^2x#

The first identity comes from dividing the equation by #a^2#, and the second by #b^2#. And remembering the basic trigonometric identities: #Soh, Cah, Toa#. 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:
#(1+tan^2x)/(1+cot^2x)=(sec^2x)/(csc^2x)=(sin^2x)/(cos^2x)=(tan^2x)^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.