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Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 25 maths textbook solution

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Answer: \sin y=C \cos x

Hint: Separate the terms of x and y and then integrate them.

Given: \cos x \cos y \frac{d y}{d x}=-\sin x \sin y

Solution: \cos x \cos y \frac{d y}{d x}=-\sin x \sin y

        \begin{aligned} &\frac{\cos y}{\sin y} d y=\frac{-\sin x}{\cos x} d x \\\\ &\cot y d y=-\tan x d x \end{aligned}

         Integrating both sides

        \begin{aligned} &\int \cot y d y=-\int \tan x d x \\\\ &\log |\sin y|=-[-\log |\cos x|]+\log C \\\\ &\log \sin y=\log \cos x+\log C \end{aligned}

        \begin{aligned} &\log \sin y=\log C(\cos x) \\\\ &\sin y=C \cos x \end{aligned}

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