Why does tan^2 y = x

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

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

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

Taking the reciprocal of both sides.

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

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

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

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

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

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

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

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

Taking the reciprocal of both sides.

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

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

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

-Dan

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