Question

Simplify the expression: tan(sin^−1(x))

Answer 1

This expression means, "calculate the tangent of the angle whose sin is x". So, we know by trigonometric identities,
\[\sin^2\alpha+\cos^2\alpha=1\]\[1+\tan^2\alpha=\frac{1}{\cos^2\alpha} \Rightarrow\tan^2\alpha=\frac{1}{1-\sin^2\alpha} -1\]
with
\[\alpha=\sin^{-1}x\]Or
\[\sin\alpha=x\]So
\[\tan^2\alpha=\frac{1}{1-x^2} -1=\frac{x^2}{1-x^2}\]\[\tan(\sin^{-1}x)=\pm\frac{x}{\sqrt{1-x^2}}\]