Answers: 1
![\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}](/tpl/images/0368/5781/2b4ae.png)
then 
is positive. So take the positive square root and you get
![\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}](/tpl/images/0368/5781/dc0fe.png)
![\bf tan\left[ sin^{-1}(x) \right]\implies \begin{cases} x=sin(\theta)\\ or\\ \cfrac{x}{1}=sin(\theta)\implies \cfrac{x=opposite=b}{1=radius=c} \end{cases}\\ \\ \quad \\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \\ \quad \\ \sqrt{1^2-x^2}=a=cos(\theta)\\ ----------------------------\\ thus \\ \quad \\ tan\left[ sin^{-1}(x) \right]\implies tan(\theta)\implies \cfrac{sin(\theta)}{cos(\theta)}\implies \cfrac{\frac{x}{1}}{\sqrt{1-x^2}} \\ \quad \\ \cfrac{x}{\sqrt{1-x^2}}](/tpl/images/0368/5781/593bb.png)
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