Question

If $\sin ({\sin }^{-1}\frac{1}{5}+{\cos }^{-1}x)=1$, then $x=$

• A. $1$
• B. $0$
• C. $\frac{4}{5}$
• D. $\frac{1}{5}$

Answer 1

Answer: (D)

$\frac{1}{5}$

Given, $\sin ({\sin }^{-1}\frac{1}{5}+{\cos }^{-1}x)=1$
$\Rightarrow \sin \left({\sin }^{-1}\frac{1}{5}+{\sin }^{-1}\sqrt{1-x^2}\right)=1$
$\Rightarrow \sin \left({\sin }^{-1}(\frac{1}{5}\sqrt{1+x^2}+x\sqrt{1-{\left(\frac{1}{5}\right)}^2})\right)=1$
$\Rightarrow \frac{\sqrt{1-x^2}}{5}+\frac{x\sqrt{24}}{5}=1\Rightarrow \frac{\sqrt{1-x^2}}{5}=1-\frac{x\sqrt{24}}{5}$
$\Rightarrow \frac{1-x^2}{25}=1+\frac{{24x}^2}{25}-\frac{2x\sqrt{24}}{5}$
$\Rightarrow 1-x^2=25+{24x}^2-10x\sqrt{24}$

$\Rightarrow x=\frac{1}{5}$