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Please solve RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question , Question 23 maths textbook solution. 

Answers (1)

Answer : (c) is the correct option

Hint :

\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|

Given : y=2 \sin ^{2} x, y=\cos 2 x

Solution :

y=2 \sin ^{2} x                                      (1)

Differentiating (1) w.r.t x, we get

\begin{aligned} &\frac{d y}{d x}=4 \sin x \cos x \\ &{\left[\frac{d y}{d x}\right]_{x=\frac{\pi}{6}}=4 \sin \frac{\pi}{6} \cos \frac{\pi}{6}=4 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}=\sqrt{3}=m_{1}} \end{aligned}

And y=\cos\; 2x                                (2)

Differentiating (2) w.r.t y, we get

\begin{aligned} &\frac{d y}{d x}=-2 \sin 2 x \\ &{\left[\frac{d y}{d x}\right]_{x=\frac{\pi}{6}}=-2 \sin \left(2 \times \frac{\pi}{6}\right)=-2 \times \sin \frac{\pi}{3}=-2 \times \frac{\sqrt{3}}{2}=-\sqrt{3}} \end{aligned}

\begin{aligned} &\tan \theta=\left|\frac{m_{1}-m_{2}}{1-3}\right| \\ &=\left|\frac{\sqrt{3}+\sqrt{3}}{-2}\right|=\left|\frac{2 \sqrt{3}}{-2}\right|=+\sqrt{3} \end{aligned}

\begin{aligned} &\tan \theta=+\sqrt{3} \\ &\theta=\frac{\pi}{3} \end{aligned}

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