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We know that $\arctan(\tan(x))=x$ when $x$ lies between $-\pi/2$ and $+\pi/2$; but do you know a way to transform the expression $\arctan(a\tan(x))$, where $a$ is a real number between $0$ and $1$?

I thought $a$ could be transformed with trigonometric functions, such as $a=\sin(\alpha)\cos(x)$, but $\arctan(\sin(\alpha)\sin(x))$ does not remind me anything.

Maybe there is no further possible transformation?

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Well, you can convert the arctan to other inverse-trig functions, such as arcsin:

$$ \arctan(a \tan(x)) = \arcsin\left(\frac{a \sin(x)}{\sqrt{\cos(x)^2 + a^2 \sin(x)^2}}\right) $$

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  • $\begingroup$ Thanks, but not easy to manipulate. By the way, I browsed WolframAlpha and found no practical alternative formula. I guess the question is hopeless... $\endgroup$ – Andrew Apr 14 at 17:14

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