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Can anyone help with this question. I'm not sure what this question meant to do.

Thanks for help.

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closed as off-topic by user223391, JonMark Perry, user296602, John B, Watson Apr 14 '16 at 8:05

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, JonMark Perry, Community, John B, Watson
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  • $\begingroup$ You can find an expression for $cos(x)$ by taking the derivative on both sides. So you need to apply the quotient rule $\endgroup$ – Walt van Amstel Mar 30 '16 at 6:39
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$$\sin(x)=\frac{2t}{(1+t^2)}$$

Hint: $$\tan(x)=\frac{\sin x}{\cos x}=\frac{\sin x}{\sqrt{1-\sin^2 x}}$$

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Let $t=\tan(y/2)$, then $\sin(x) = \frac{2t}{1+t^2} = \sin(y)$ and $\frac{1-t^2}{1+t^2} = \cos(y)$, which means that $$\tan(x)=\tan(y) =\frac{\sin(y)}{\cos(y)}=\frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}=\frac{2t}{1-t^2}.$$

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This question asks to find $\tan(x)$ in the form of "t". The domain is [$0$,$\pi/2$] where all are positive.

We know $\cos x$= $\sqrt{1-\sin^2x}$

Put value of $\sin(x)$ in the given expression to get $\cos(x)$ in the form of "t".

Then find $\tan(x)$ = $\frac{\sin(x)}{\cos(x)}$ in the form of "t".

Final come out will be $\tan(x)$=$\frac{2t}{1-t^2}.$

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