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Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 22

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Answer: \tan x \times \sin x=c

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

Given: \tan y d x+\sec ^{2} y \tan x d y=0

Solution: \tan y d x+\sec ^{2} y \tan x d y=0

        \begin{aligned} &\Rightarrow \sec ^{2} y \tan x d y=-\tan y d x \\\\ &\Rightarrow \frac{\sec ^{2} y}{\tan y} d y=-\frac{1}{\tan x} d x \\\\ &\Rightarrow \frac{1}{\cos ^{2} y} \times \frac{\cos y}{\sin y} d y=-\cot x d x \end{aligned}

        \begin{aligned} &\Rightarrow \frac{1}{\sin y \cos y} d y=-\cot x d x \\\\ &\Rightarrow \frac{2}{\sin 2 y}=-\cot x d x \end{aligned}

          Integrating both sides

        \begin{aligned} &\Rightarrow 2 \int \operatorname{cosec} 2 y d y=-\int \cot x d x \\\\ &\Rightarrow \log \tan x=-\log \sin x+\log c \\\\ &\Rightarrow \log \tan x+\log \sin x=\log c \\\\ &\Rightarrow \log (\tan x \times \sin x)=\log c \\\\ &\Rightarrow \tan x \times \sin x=c \end{aligned}

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