Question

The sum of all the values of $\theta (0\le \theta \le 2\pi )$ which satisfies both the equations $r\sin \theta =3$ and $r=4(1+\sin \theta )$ is

• A. $\frac{\pi }{6}$
• B. $\frac{5\pi }{6}$
• C. $\pi$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $\pi$

$r\sin \theta =3\cdots \left(1\right)$
$r=4(1+\sin \theta )\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$, we get
$\frac{3}{\sin \theta }=4(1+\sin \theta )\Rightarrow 4{\sin }^2\theta +4\sin \theta -3=0\Rightarrow (2\sin \theta +3)(2\sin \theta -1)=0\Rightarrow \sin \theta =\frac{1}{2}(\because \sin \theta \in [-1,1\left]\right)\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6}(\because \theta \in [0,2\pi \left]\right)$

Hence, the sum of all values of $\theta$ is $\pi$.