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Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 47

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

Hint: To solve this equation we use uv'  form

Given:  x \sin (a+y)+\sin a \cdot \cos (a+y)=0

Solution:   (u v)^{\prime}=u^{\prime} v+v^{\prime} u

        \sin (a+y)+x \cdot \cos (a+y) y^{\prime}+\sin a\left(-\sin (a+y)\left(0+y^{\prime}\right)=0\right.

        y^{\prime}[x \cdot \cos (a+y)+(-\sin a) \cdot \sin (a+y)]=-\sin (a+y)

        y^{\prime}=\frac{-\sin (a+y)}{x \cdot \cos (a+y)+(-\sin a) \cdot(\sin a+y)} \cdot \frac{\sin (a+y)}{\sin (a+y)}

        \frac{d y}{d x}=\frac{-\sin ^{2}(a+y)}{-\sin a\left(\cos ^{2}(a+y)+\sin ^{2}(a+y)\right.} \quad\left(\sin ^{2} x+\cos ^{2} x=1\right)

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

Hence proved

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