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Provide solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 64 Subquestion (i) textbook solution.

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Answer : y=2 \tan \frac{x}{2}-x+c

Hint                : You must know the rules of solving differential equation and integration

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

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

                         \begin{aligned} &\frac{d y}{d x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} \\ &\frac{d y}{d x}=2 \tan ^{2} \frac{x}{2} \end{aligned}

                         \begin{aligned} &d y=\left(\tan ^{2} \frac{x}{2}\right) d x \\ &d y=\left(\sec ^{2} \frac{x}{2}-1\right) d x \end{aligned}

Integrating both sides,

                    \begin{gathered} \int d y=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x \\ \int d y=\int \sec ^{2} \frac{x}{2}-\int 1 d x \\ y=2 \tan \frac{x}{2}-x+c \end{gathered}

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