write the given expression as an algebraic expression in x. tan(2 sin^{-1}(x))

Explanation:

Step1: Let $\theta=\sin^{-1}(x)$

So, $\sin\theta = x$, and by the Pythagorean identity $\cos\theta=\sqrt{1 - x^{2}}$ (assuming $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ where $\cos\theta\geq0$)

Step2: Use the double - angle formula for tangent

The double - angle formula for tangent is $\tan(2\theta)=\frac{2\tan\theta}{1-\tan^{2}\theta}$ Since $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{x}{\sqrt{1 - x^{2}}}$

Step3: Substitute $\tan\theta$ into the double - angle formula

[ \begin{align*} \tan(2\sin^{-1}(x))&=\frac{2\frac{x}{\sqrt{1 - x^{2}}}}{1-\left(\frac{x}{\sqrt{1 - x^{2}}}\right)^{2}}\ &=\frac{\frac{2x}{\sqrt{1 - x^{2}}}}{\frac{1 - x^{2}-x^{2}}{1 - x^{2}}}\ &=\frac{2x}{\sqrt{1 - x^{2}}}\times\frac{1 - x^{2}}{1 - 2x^{2}}\ &=\frac{2x\sqrt{1 - x^{2}}}{1 - 2x^{2}} \end{align*} ]

Answer:

$\frac{2x\sqrt{1 - x^{2}}}{1 - 2x^{2}}$