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

2 Answers 2

up vote 4 down vote accepted

Yes, you did good. The denominator can be simplified to $|x-1|$.

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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
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Look at the denominator of your answer. Look at the denominator in the option. What do you see? –  Gerry Myerson May 6 '11 at 0:00
    
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*}

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@Matt: Hope this solution helps. –  user9413 May 8 '11 at 14:56

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