Question

If $\sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\ldots \infty\right)$
$+\cos ^{-1}\left(x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4}-\ldots \infty\right)=\frac{\pi}{2}$ and
$0 < x < \sqrt{2}$, then $x$ is equal to

• A. $\frac{1}{2}$
• B. 1
• C. $-\frac{1}{2}$
• D. $-1$

Answer 1

Answer: (B)

Given that, $\sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\ldots \infty\right)$
$+\cos ^{-1}\left(x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4}-\ldots \infty\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}\left[\frac{x}{1-\frac{x}{2}}\right]+\cos ^{-1}\left[\frac{x^{2}}{1-\frac{x^{2}}{2}}\right]=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}\left[\frac{2 x}{2-x}\right]+\cos ^{-1}\left[\frac{2 x^{2}}{2-x^{2}}\right]=\frac{\pi}{2}$
$\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$
$\Rightarrow \frac{2 x}{2-x}=\frac{2 x^{2}}{2-x^{2}}$
$\Rightarrow \left(2-x^{2}\right)=x(2-x)$
$\Rightarrow 2-x^{2}=2 x-x^{2}$
$\Rightarrow 2 \,x=2$
$\Rightarrow x=1$