Question

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

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

Answer 1

The correct option is B

$1$

Explanation for the correct option.

Step 1: Simplify the equation

Given that, ${\sin }^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\dots \infty \right)+{\cos }^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\dots \infty \right)=\frac{\pi }{2}$.

Recall that sum of an infinite GP is $\frac{a}{1-r}$ for $\left|r\right|<1$.
So $x-\frac{x^2}{2}+\frac{x^3}{4}-\cdots \infty =\frac{x}{1+\frac{x}{2}}$
And also $x^2-\frac{x^2}{2}+\frac{x^6}{4}-\cdots \infty =\frac{x^2}{1+\frac{x^2}{2}}$

Step 2: Solve equation for $x$

$$\begin{gathered} \Rightarrow {\sin }^{-1}\left(\frac{x}{1+\frac{x}{2}}\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\frac{x^2}{1+\frac{x^2}{2}}\right) \\ \Rightarrow {\sin }^{-1}\left(\frac{2x}{x+2}\right)={\sin }^{-1}\left(\frac{2x^2}{x^2+2}\right)\left[\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}}\right] \\ \Rightarrow \frac{1}{x+2}=\frac{x}{x^2+2} \\ \Rightarrow x^2+2x=x^2+2 \\ \Rightarrow 2x=2 \\ \Rightarrow x=1 \end{gathered}$$.

So, the value of $x=1$.

Hence option B is correct.