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The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$ is
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The correct answer is:
$-\frac{2 \sqrt{6}}{5}$
$\cos \left[2 \cos ^{-1} x+\sin ^{-1} x\right]$
$=\cos \left[\cos ^{-1} x+\cos ^{-1} x+\sin ^{-1} x\right]$
$=\cos \left[\cos ^{-1} x+\frac{\pi}{2}\right]=-\sin \left[\cos ^{-1} x\right]$
$=-\sin \left[\sin ^{-1} \sqrt{1-x^{2}}\right]=-\sqrt{1-x^{2}}$
$=-\sqrt{1-\left(\frac{1}{5}\right)^{2}}=-\sqrt{\frac{24}{25}}=-\frac{2 \sqrt{6}}{5}$
$=\cos \left[\cos ^{-1} x+\cos ^{-1} x+\sin ^{-1} x\right]$
$=\cos \left[\cos ^{-1} x+\frac{\pi}{2}\right]=-\sin \left[\cos ^{-1} x\right]$
$=-\sin \left[\sin ^{-1} \sqrt{1-x^{2}}\right]=-\sqrt{1-x^{2}}$
$=-\sqrt{1-\left(\frac{1}{5}\right)^{2}}=-\sqrt{\frac{24}{25}}=-\frac{2 \sqrt{6}}{5}$
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