# $\frac{x_1^2}{\cos^2 \left( \arctan \frac{x_2}{x_1} \right)} =x_1^2 +x_2^2$?

Is the relationship true? I don't seem able to prove it myself. I know that $$\cos(\arctan(x)) =\frac{1}{\sqrt{1+x^2}}$$ but I am at a loss here.

Thank you.

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## 1 Answer

So, $$\cos\left(\arctan \frac{x_2}{x_1}\right)=\frac1{\sqrt{1+\left(\frac{x_2}{x_1}\right)^2}}$$

$$\implies\cos^2\left(\arctan \frac{x_2}{x_1}\right)=\frac1{1+\left(\frac{x_2}{x_1}\right)^2}=\frac{x_1^2}{x_1^2+x_2^2}$$

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@AhaanRungta It's not anything that I did not post above, yet it is helpful because I have not worked with trig functions for a while. What is your problem? –  JohnK Nov 25 '13 at 16:51
Never mind, I misread. Sorry for that. I am deleting my comment. You should too. –  Ahaan S. Rungta Nov 25 '13 at 17:04