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  • Please solve RD Sharma class 12 chapter 10 Differentiation exercise Multiple choice question 5 maths textbook solution

Please solve RD Sharma class 12 chapter 10 Differentiation exercise Multiple choice question 5 maths textbook solution

Answers (1)

best_answer

Answer:

        \frac{1}{2}

Hint:

        Differentiate the function and replace x by \frac{\pi }{6} solve

Given:

        f(x)=\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}, 0 \leq x \leq \frac{\pi}{2}, f^{\prime}\left(\frac{\pi}{6}\right)

Solution:  

        y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)

            =\tan ^{-1}\left(\sqrt{\frac{1-\cos \left(\frac{\pi}{2}+x\right)}{1+\cos \left(\frac{\pi}{2}+x\right)}}\right)

            =\tan ^{-1}\left(\sqrt{\frac{2 \sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \cos ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}}\right)

        y=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right)=\frac{\pi}{4}+\frac{x}{2} \; \; \; \; \; \; \; \quad\left[\tan ^{-1}(\tan x)=x\right]

        \frac{d y}{d x}=\frac{1}{2}

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