For 0≤cos-1x≤π and -π2≤sin-1x≤π2 then value of cossin-1x+2cos-1x at x=1/5 is

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For $0 \leq \cos^{- 1} x \leq π$ and $\frac{- π}{2} \leq \sin^{- 1} x \leq \frac{π}{2}$ then value of $\cos (\sin^{- 1} x + 2 \cos^{- 1} x)$ at $x = 1 / 5$ is

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$\frac{- 2 \sqrt{6}}{5}$

$\frac{- \sqrt{6}}{5}$

$\frac{2 \sqrt{6}}{5}$

$\frac{\sqrt{6}}{5}$

answer is A.

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Detailed Solution

$\cos (\sin^{- 1} x + 2 \cos^{- 1} x) = \cos (\frac{π}{2} + \cos^{- 1} x) = - \sin (\cos^{- 1} x) (let \cos^{- 1} x = θ \Rightarrow x = \cos θ)$
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