# Given $\sin x =\frac{2t}{(1+t^2)}$, find $\tan x$ if $x \in [0, \pi/2]$ [closed]

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

Thanks for help.

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

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• You can find an expression for $cos(x)$ by taking the derivative on both sides. So you need to apply the quotient rule – Walt van Amstel Mar 30 '16 at 6:39

$$\sin(x)=\frac{2t}{(1+t^2)}$$

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

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 $$\tan(x)=\tan(y) =\frac{\sin(y)}{\cos(y)}=\frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}=\frac{2t}{1-t^2}.$$

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 $\tan(x)$=$\frac{2t}{1-t^2}.$