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Show that $ \tan(\arcsin x)= \frac{x}{\sqrt{1-x^2}} $ I don't know where to start.

Martin Sleziak
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Lyndt
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    For $0\lt x\lt 1$, draw a right triangle with hypotenuse $1$ and one of the sides equal to $x$. Then take care of other values of $x$, namely $x=1$, $0$, and $-1\le x\lt 0$. – André Nicolas Sep 30 '15 at 02:58
  • See also: http://math.stackexchange.com/questions/368603/how-do-i-write-a-trig-function-that-includes-inverses-in-terms-of-another-variab – Martin Sleziak Oct 19 '15 at 13:17
  • @Harish Why was this tagged (proof-verification)? If you read the tag-info, you will see that this tag is supposed to be used if the OP wrote down a proof of some fact and they want to ask whether the proof is correct. But the OP did not post any proof of the equality in the question. – Martin Sleziak Oct 19 '15 at 13:25

2 Answers2

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Notice, let $$\sin^{-1}(x)=\theta \implies x=\sin \theta $$ $\forall\ \ 0<\theta<\pi/2 \implies \ \sin\theta>0, \ \ \cos \theta>0$

Now, we know that $$\tan\theta=\frac{\sin \theta}{\cos\theta}$$ $$=\frac{\sin \theta}{\sqrt{1-\sin^2\theta}}$$ Now, setting $x=\sin\theta $ $$\tan\theta =\frac{x}{\sqrt{1-x^2}}$$ Now, setting $\theta=\sin^{-1}(x)$ , we get $$\tan(\sin^{-1}(x)) =\frac{x}{\sqrt{1-x^2}}$$

Harish Chandra Rajpoot
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Hint: Setting $\sin^{-1} x = t$, for $|t| \leq \frac{\pi}{2}$, then we get $x = \sin t$ and $\frac{1}{\tan^2 t}+1 = \frac{1}{\sin^2 t}$.

GAVD
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