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Question

Mathematics Question on Properties of Inverse Trigonometric Functions

The value of cos(2cos1x+sin1x)\cos\left( 2 \cos^{-1} x + \sin^{-1} x\right) at x=15x = \frac{1}{5} is

A

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

B

26 - 2 \sqrt{6}

C

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

D

None

Answer

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

Explanation

Solution

cos(2cos1x+sin1x)\cos\left(2 \cos^{-1} x + \sin^{-1}x \right) =cos(cos1x+cos1x+sin1x)= \cos\left(\cos^{-1} x + \cos^{-1} x + \sin^{-1}x\right) =cos(cos1x+π/2)=sincos1x= \cos\left(\cos^{-1} x + \pi /2\right) =- \sin\cos^{-1} x =sinsin11x2=1x2= - \sin\sin^{-1} \sqrt{1-x^{2} } = - \sqrt{1-x^{2}} =1(15)2=2425=265= - \sqrt{1- \left(\frac{1}{5}\right)^{2}} = - \sqrt{\frac{24}{25}} = - \frac{2\sqrt{6}}{5}