Show that $ \tan(\arcsin x)= \frac{x}{\sqrt{1-x^2}} $
Show that $ \tan(\arcsin x)= \frac{x}{\sqrt{1-x^2}} $ I don't know where to start.
$\endgroup$ 32 Answers
$\begingroup$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 $$\tan(\sin^{-1}(x)) =\frac{x}{\sqrt{1-x^2}}$$
$\endgroup$ 7 $\begingroup$Hint: Setting $\sin^{-1} x = t$, for $|t| \leq \frac{\pi}{2}$, then we get $x = \sin t$ and $\frac{1}{\tan^2 t}+1 = \frac{1}{\sin^2 t}$.
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