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Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions Question 26 Maths Textbbok Solution.

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Answer: \frac{1}{\sqrt{2}}<x \leq 1

Hint: Substitute range values to the function.

Given: \cos ^{-1} x>\sin ^{-1} x


            \cos ^{-1} x>\sin ^{-1} x

            \cos ^{-1} x>\frac{\pi}{2}-\cos ^{-1} x

            2 \cos ^{-1} x>\frac{\pi}{2}

            \cos ^{-1} x>\frac{\pi}{4}

            x>\cos \frac{\pi}{4}


           \therefore We know that the maximum value of cosine function is 1.

           \therefore \frac{1}{\sqrt{2}}<x \leq 1


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