Question

$\sin^{- 1}\left( x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - ...... \right) + \cos^{- 1}\left( x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - ...... \right) = \frac{\pi}{2}$ for

0<|x|<$\sqrt{2}$, then x equals

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

Answer 1

Answer: (B)

1

Answer 2

The given equation implies

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

⇒ $\frac{2x}{2 + x} = \frac{2x^{2}}{2 + x^{2}} \Rightarrow x = 1$