Question

If ${\sin }^{-1}$ $[x-\frac{x^2}{2}+\frac{x^3}{4}-.......\infty ]$ +${\cos }^{-1}$ $[x^2-\frac{x^4}{2}+\frac{x^6}{4}-.......\infty ]$=$\frac{\pi }{2}$ and 0<x$\le \sqrt{2}$ then $x=$

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

Answer 1

The correct option is B $1$

As we know that

Sum of an infinite $GP$ is $\frac{a}{1-r}$ for $|r|<1$

So $x-\frac{x^2}{2}+\frac{x^3}{4}-\cdots \infty =\frac{x}{1+\frac{x}{2}}$

And also $x^2-\frac{x^2}{2}+\frac{x^6}{4}-\cdots \infty =\frac{x^2}{1+\frac{x^2}{2}}$

So ${\text{sin}}^{-1}(x-\frac{x^2}{2}+\frac{x^3}{4}-\cdots \infty )+{\text{cos}}^{-1}(x^2-\frac{x^4}{4}+\frac{x^6}{4}-\cdots \infty )=\frac{\pi }{2}$

$⟹{\text{sin}}^{-1}(\frac{x}{1+\frac{x}{2}})=\frac{\pi }{2}-{\text{cos}}^{-1}(\frac{x^2}{1+\frac{x^2}{2}})$

$⟹{\text{sin}}^{-1}(\frac{2x}{x+2})={\text{sin}}^{-1}(\frac{2x^2}{x^2+2})$

$⟹\frac{1}{x+2}=\frac{x}{x^2+2}$

$⟹x^2+2x=x^2+2$

$⟹2x=2$

$⟹x=1$