Hello,

I continue to struggle with trig.

The question is:

Given that $\displaystyle x = \tan y$ show that $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{1 + x^2}$.

The answer given, most of which I follow, is:

$\displaystyle \dfrac{d}{dy} (\tan y) = \sec^2 y$

Taking the reciprocal of both sides.

$\displaystyle \dfrac{dy}{dx} = \dfrac{1}{sec^2 y}$

$\displaystyle \dfrac{dy}{dx} = \dfrac{1}{1 + \tan^2 y}$

Hence, and this is the bit I don't get.

$\displaystyle \dfrac{dy}{dx} = \dfrac{1}{1 + x^2}$

It's bound to be something I'm supposed to know, something basic, but why is $\displaystyle \tan^2 y = x^2$

Thanks