Question

$\frac{2rsin\alpha }{1+2rcos\alpha }$ is equal to

• A. ${\tan }^2\theta$
• B. ${\cot }^2\theta$
• C. $\cot 2\theta$
• D. $\tan 2\theta$

Answer 1

Answer: (D)

$\tan 2\theta$

In $△OLC$

$OL=r\cos (\alpha -3\theta ),CL=r\sin (\alpha -3\theta )$

Gives

$\frac{{\sin }^3\theta }{CL}=\frac{{\cos }^3\theta }{OL}=1\Rightarrow \frac{{\sin }^3\theta }{r\sin (\alpha -3\theta )}=\frac{{\cos }^3\theta }{r\cos (\alpha -3\theta )}=1$

$\Rightarrow \frac{{\sin }^3\theta }{\sin (\alpha -3\theta )}=\frac{{\cos }^3\theta }{\cos (\alpha -3\theta )}=r$

${\cos }^4\theta -{\sin }^4\theta =\cos \theta \left(r\cos \right(\alpha -3\theta )-\sin \theta (r\sin (\alpha -3\theta ))$

$\Rightarrow \cos 2\theta =r\cos (\alpha -2\theta )\Rightarrow \cos 2\theta =r(\cos \alpha \cos 2\theta +\sin \alpha \sin 2\theta )$

$\Rightarrow (1-r\cos \alpha )\cos 2\theta =r\sin \alpha \sin 2\theta \Rightarrow \frac{1-r\cos \alpha }{r\sin \alpha }=\tan 2\theta .....\left(1\right)$

And

${\cos }^3\theta \sin \theta +{\sin }^3\theta \cos \theta =r\left(\cos \right(\alpha -3\theta )\sin \theta +\sin (\alpha -3\theta \left)\cos \theta \right)$

$\Rightarrow \sin \theta \cos \theta =r\sin (\alpha -2\theta )\Rightarrow \sin 2\theta =2r(\sin \alpha \cos 2\theta -\cos \alpha \sin 2\theta )$

$\Rightarrow (1+2r\cos \alpha )\sin 2\theta =2r\sin \alpha \cos 2\Rightarrow \frac{1+2r\cos \alpha }{2r\sin \alpha }=\cot 2\theta$

From $\left(1\right)$ and $\left(2\right)$

$\frac{1-r\cos \alpha }{r\sin \alpha }=\frac{2r\sin \alpha }{1+2r\cos \alpha }=\tan 2\theta$