# Why does tan^2 y = x

• Aug 28th 2012, 04:06 PM
Furyan
Why does tan^2 y = x^2
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
• Aug 28th 2012, 04:34 PM
topsquark
Re: Why does tan^2 y = x
Quote:

Originally Posted by Furyan
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}\$

You have been given \$\displaystyle x = tan(y)\$. Then by squaring both sides: \$\displaystyle tan^2(y) = x^2\$.

-Dan