Question

If $x=\frac{2}{3}$, then ${\sin }^2\left({\tan }^{-1}x\right)+{\cos }^2\left({\sin }^{-1}x\right)$ is equal to

• A. $\frac{99}{101}$
• B. $\frac{107}{117}$
• C. $\frac{101}{117}$
• D. none of these

Answer 1

The correct option is C $\frac{101}{117}$

Given,

$\begin{array}{c}x=\frac{2}{3}\\ Now,{\sin }^2\left({\tan }^{-1}x\right)+{\cos }^2\left({\sin }^{-1}x\right)\\ ={\left(\sin \left({\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right)\right)}^2+{\left(\cos \left({\cos }^{-1}\sqrt{1-x^2}\right)\right)}^2\\ ={\left[\frac{x^2}{1+x^2}+1-x^2\right]}_{x=\frac{2}{3}}\\ \frac{\frac{4}{9}}{1+\frac{4}{9}}+1-\frac{4}{9}=\frac{4}{13}+\frac{5}{9}=\frac{36+65}{117}=\frac{101}{117}\end{array}$