Question

lf the point $(1+\cos \theta ,\sin \theta )$ lies between the region corresponding to the acute angle between $x-3y=0$ and $x-6y=0,$then $\theta \in$

• A. $(2n\pi +2{\tan }^{-1}\frac{1}{6},2n\pi +2{\tan }^{-1}\frac{1}{3})$
• B. $(n\pi +2{\tan }^{-1}\frac{1}{6},n\pi +2{\tan }^{-1}\frac{1}{3})$
• C. $(2n\pi -2{\tan }^{-1}\frac{1}{3},2n\pi -2{\tan }^{-1}\frac{1}{6})$
• D. None of these

Answer 1

The correct option is A $(2n\pi +2{\tan }^{-1}\frac{1}{6},2n\pi +2{\tan }^{-1}\frac{1}{3})$

The slope of OP $=\frac{\sin \theta }{1+\cos \theta }=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}=\tan \left(\frac{\theta }{2}\right)$
Slope of line $x-3y=0$ is $\frac{1}{3}$
Slope of line $x-6y=0$ is $\frac{1}{6}$
Since, P lies in the region of the acute angle between the lines $\tan \frac{\theta }{2}>\frac{1}{6}$and $\tan \frac{\theta }{2}<\frac{1}{3}$
$\Rightarrow \frac{\theta }{2}>{\tan }^{-1}\frac{1}{6}$ and $\frac{\theta }{2}<{\tan }^{-1}\frac{1}{3}$
i.e., $2n\pi +2{\tan }^{-1}\frac{1}{6}<\theta <2n\pi +2{\tan }^{-1}\frac{1}{3}$