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Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise Multiple Choice Questions Question 6 Maths Textbook Solution.

Answers (1)

Answer: \frac{\sqrt{3}}{2}

Hint: Take sin and cos relative formula like \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}

We can simply calculate the variable.

Given:  \sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}

Solution:

            \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}

           \therefore \sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}

            \begin{aligned} \frac{\pi}{2}-\cos ^{-1} x &-\cos ^{-1} x=\frac{\pi}{6} \\ -2 \cos ^{-1} x &=\frac{\pi}{6}-\frac{\pi}{2} \\ -2 \cos ^{-1} x &=\frac{-\pi}{3} \\ \cos ^{-1} x &=\frac{\pi}{6} \\ x &=\cos \frac{\pi}{6} \\ x &=\frac{\sqrt{3}}{2} \end{aligned}

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