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  • Provide Solution for RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 2

Provide Solution for RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 2

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Answer:

            \frac{d y}{d x}=\tan \left(\frac{\theta}{2}\right)

Hint:

            \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}

Given:

            x=a(\theta+\sin x) \ , y=a(1-\cos \theta) \\

Solution:

x=a(\theta+\sin x) \

\ \frac{d x}{d \theta}=a(1+\cos \theta) \\

y=a(1-\cos \theta) \\

\frac{d y}{d \theta}=a(-\sin \theta)=a \sin \theta \\

\begin{aligned} & &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{dx}{d\theta}}=\frac{a \sin \theta}{a(1+\cos \theta)} \end{aligned}

\frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}

\begin{aligned} & \\ &=\frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \end{aligned}                                                                                                \left[\begin{array}{l} \sin 2 \theta=2 \sin \theta \cos \theta \\ 1+\cos 2 \theta=2 \cos ^{2} \theta \end{array}\right]

=\frac{\sin \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)}                                                                                                      \left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]

\frac{d y}{d x}=\tan \left(\frac{\theta}{2}\right)

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