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Question:

$\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$

Solution:

The given differential equation is:

$\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$

$\Rightarrow \frac{d y}{d x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=\tan ^{2} \frac{x}{2}$

$\Rightarrow \frac{d y}{d x}=\left(\sec ^{2} \frac{x}{2}-1\right)$

Separating the variables,we get:

$d y=\left(\sec ^{2} \frac{x}{2}-1\right) d x$

Now, integrating both sides of this equation, we get:

$\int d y=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x=\int \sec ^{2} \frac{x}{2} d x-\int d x$

$\Rightarrow y=2 \tan \frac{x}{2}-x+\mathrm{C}$

This is the required general solution of the given differential equation.