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
