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 $0<x<\sqrt{2}$ then x =

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

Answer 1

The correct option is B $1$

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

As we know that

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

So $x-\frac{x^2}{2}+\frac{x^3}{4}+..............\infty =$ $x^2-\frac{x^4}{2}+\frac{x^6}{4}-...........\infty$

As we know that

Sum of infinite $GP$ i.e $a+ar+a{r}^2+\cdots \infty =\frac{a}{1-r}$

$⟹\frac{x}{1+\frac{x}{2}}=\frac{x^2}{1+\frac{x^2}{2}}$

$⟹\frac{1}{x}+\frac{1}{2}=\frac{1}{x^2}+\frac{1}{2}$

$⟹x=x^2⟹x=1$