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  • Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Multiple choice question 15

Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Multiple choice question 15

Answers (1)

best_answer

Answer:

        -\frac{1}{2}

Hint:

        Differentiate the function w.r.t x

Given:

        \frac{d}{d x}\left[\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\right]

Solution:  

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

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

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

        =\tan ^{-1}\left[\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right]

        u=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}-\tan \frac{x}{2}}{1+\tan \frac{\pi}{4} \times \tan \frac{x}{2}}\right]

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

        \begin{aligned} &u=\frac{\pi}{4}-\frac{x}{2} \\\\ &\frac{d u}{d x}=0-\frac{1}{2} \\\\ &\frac{d u}{d x}=-\frac{1}{2} \end{aligned}

 

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