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Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 57 maths textbook solution

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Answer: \frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}

Hint: To solve this equation we make statement  \frac{\cos (a+y)-a}{\cos (a+y)}=x

Given: \cos y=x \cdot \cos (a+y)

Solution:  we have

        \cos y=x \cdot \cos (a+y)

        \frac{\cos ((a+y)-a)}{\cos (a+y)}=x

        \frac{\cos (a+y) \cos a+\sin (a+y) \sin a}{\cos (a+y)}=x

        \frac{\cos (a+y) \cos a}{\cos (a+y)}+\frac{\sin (a+y) \cdot \sin a}{\cos (a+y)}

        \frac{d}{d x}(\cos a+\tan (a+y) \sin a)=\frac{d x}{d x}

        \sin a \cdot \frac{d}{d x}(\tan (a+y))=1 \quad\left[\because \frac{d}{d x} \tan x=\sec ^{2} x\right]

        \sin a \cdot \sec ^{2}(a+y) \frac{d}{d x}(a+y)=1

        \begin{aligned} &\frac{\sin a}{\cos ^{2}(a+y)} \cdot \frac{d y}{d x}=1 \\\\ &\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a} \end{aligned}

Hence proved


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