# If $\cos^{-1}x=\tan^{-1}x,\;show\;that\;\sin(\cos^{-1}x)=x^2$

Toolbox:
• $$cos^{-1}x = tan^{-1} \large\frac{\sqrt{1-x^2}}{x}$$
• $$sincos^{-1}x=\sqrt{1-x^2}$$
given $$cos^{-1}x=tan^{-1}x$$

By substituting the above formula of $$cos^{-1}x$$
$$\Rightarrow tan^{-1} \large\frac{\sqrt{1-x^2}}{x}=tan^{-1}x$$
$$\Rightarrow \large\frac{ \sqrt{1-x^2}}{x}=x$$
$$\Rightarrow \sqrt{1-x^2}=x^2$$

Now $$sin\: cos^{-1}x = \sqrt{1-x^2}=x^2$$

edited Mar 19, 2013