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Question
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Solution
We have,
\[\cos x \cos y \frac{dy}{dx} = - \sin x \sin y \]
\[ \Rightarrow \frac{\cos y}{\sin y}dy = \frac{- \sin x}{\cos x}dx\]
\[ \Rightarrow \cot y\ dy = - \tan x\ dx\]
Integrating both sides, we get
\[\int \cot y\ dy = - \int\tan x\ dx\]
\[ \Rightarrow \log \left| \sin y \right| = - \log \left| \sec x \right| + \log C\]
\[ \Rightarrow \log \left| \sin y \right| = \log \left| \cos x \right| + \log C\]
\[ \Rightarrow \sin y = C \cos x\]
\[\text{ Hence, }\sin y = C \cos x\text{ is the required solution . }\]
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