# Simple Trig Question

Hi I am not sure I am solving this trig question correctly:

$\tan\left(\sin^{-1}\left(\dfrac{2\sqrt{x}}{1+x}\right)\right) = ?$

I drew a right triangle and set an angle equal to $\sin^{-1}\left(\dfrac{2\sqrt{x}}{1+x}\right)$

so that I could find the third (adjacent) side after setting the opposite as $2\sqrt{x}$ and the hypotenuse as $1+x$ by using the Pythagorean theorem.

I believe after using the Pythagorean theorem and doing the algebra that the adjacent side is $\sqrt{x^2-2x+1}$

So would it then follow that the answer to the question would be $\dfrac{2\sqrt{x}}{\sqrt{x^2-2x+1}}$?

thanks

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Alternatively, you can rewrite $\tan\;x$ as $\frac{\sin\;x}{\sqrt{1-\sin^2 x}}$... – J. M. May 5 '11 at 23:58

Yes, you did good. The denominator can be simplified to $|x-1|$.
It is a multiple choice question and the answer I got wasn't an option but $\dfrac{2\sqrt{x}}{x-1}$ is. how can it be simplified? by rationailizing? – Matt May 5 '11 at 23:58
ahhh, $(x-1)^2$, thanks! – Matt May 6 '11 at 0:07
@Matt: Take, for example, $x=1/4$. Then your answer is right but the one that comes from $\frac{2\sqrt{x}}{x-1}$ is wrong. – André Nicolas May 6 '11 at 0:23
Another way of doing this by substitution. Note that $\sin{2x} =\frac{2 \tan{x}}{1+\tan^{2}{x}}$. So what I did is to put $x = \tan^{2}{t}$. Then what you want is precisely $\tan{2t}$. Which you can obtain by the formula \begin{align*} \tan{2t} &=\frac{2 \cdot \tan{t}}{1 - \tan^{2}{t}} \\ &= \frac{2 \sqrt{x}}{1-x} \qquad \Bigl[ \text{since} \ x=\tan^{2}{t} \ \Bigr] \end{align*}