If sin - 1 ( x - x 2 2 + x 3 4 - ? ) + cos - 1 ( x 2 - x 4 2 + x 6 4 - ? ) = 2 for 0 | x | 2 , then x equals [2001]

Question

If $\sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots) + \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < | x | < \sqrt{2}$ , then $x$ equals [2001]

Smart Achievers · Accepted Answer

(2)

$\sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots) + \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2}$

$\Rightarrow \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2} - \sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots)$

$\Rightarrow \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \cos^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots)$

$\Rightarrow x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots = x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots$

On both sides we have G.P. infinite terms.

$∴$ $\frac{x^{2}}{1 - (- \frac{x^{2}}{2})} = \frac{x}{1 - (- \frac{x}{2})}$ $\Rightarrow \frac{2 x^{2}}{2 + x^{2}} = \frac{2 x}{2 + x}$

$\Rightarrow 2 x + x^{3} = 2 x^{2} + x^{3}$ $\Rightarrow x (x - 1) = 0$

$\Rightarrow x = 0, 1$ but $0 < | x | < \sqrt{2} \Rightarrow x = 1$

Solution

Q.
If $\sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots) + \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < | x | < \sqrt{2}$ , then $x$ equals [2001]

1 $\frac{1}{2}$

2 $1$

3 $- \frac{1}{2}$

4 $- 1$

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Solution

Q. If sin-1(x-x22+x34-⋯)+cos-1(x2-x42+x64-⋯)=π2 for 0<|x|<2, then x equals [2001]

1 12

2 1

3 -12