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#### Please solve RD Sharma class 12 chapter 10 Differentiation exercise Multiple choice question 9 maths textbook solution

$|\sec \theta|$

Hint:

Differentiate x and y w.r.t $\theta$, then divide and solve

Given:

$x=a \cos ^{3} \theta, y=a \sin ^{3} \theta, \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=$

Solution:

\begin{aligned} &x=a \cos ^{3} \theta \\\\ &\frac{d x}{d \theta}=a \frac{d}{d \theta}\left(\cos ^{3} \theta\right) \\\\ &\frac{d x}{d \theta}=a(3) \cos ^{2} \frac{d}{d \theta}(\cos \theta) \end{aligned}

$\frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta$           ....................(1)

\begin{aligned} &y=a \sin \theta \\\\ &\frac{d y}{d \theta}=a \frac{d}{d \theta}\left(\sin ^{3} \theta\right) \end{aligned}

$=3 a \sin ^{2} \theta \frac{d}{d \theta}(\sin \theta)$

$\frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta$        ......................(2)

Dividing (2) by (1) we get

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}$

\begin{aligned} &\frac{d y}{d x}=\frac{\sin \theta}{-\cos \theta}=-\tan \theta \\\\ &\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=\sqrt{1+\tan ^{2} \theta}=\sqrt{\sec ^{2} \theta}=|\sec \theta| \end{aligned}