Question

If $\frac{1}{\sqrt{2}}$< x < 1 then cos^–1x+cos^–1$\left( \frac{x + \sqrt{1 - x^{2}}}{\sqrt{2}} \right)$is equal to –

• A. 2 cos^–1x –$\frac{\pi}{4}$
• B. 2 cos–1x
• C. $\frac{\pi}{4}$
• D. 0

Answer 1

Answer: (C)

$\frac{\pi}{4}$

Answer 2

Here, the expression could be written as

̃ cos^–1x +cos^–1 $\left\{ x.\frac{1}{\sqrt{2}} + \sqrt{1 - x^{2}}.\sqrt{1 - \left( \frac{1}{\sqrt{2}} \right)^{2}} \right\}$

̃ cos^–1 x + cos^–1 $\frac{1}{\sqrt{2}}$– cos^–1x

$\left\{ \because\frac{1}{\sqrt{2}} < x \Rightarrow \cos^{- 1}\frac{1}{\sqrt{2}} > \cos^{- 1}x \right\}$

̃ cos–1$\frac{1}{\sqrt{2}}$=$\frac{\pi}{4}$