Question

The number of solutions of the equation
${\text{cosec}}^{-1}(\sin \theta +4{\sin }^2\theta )+{\sec }^{-1}(-1+6\sin \theta )=\frac{\pi }{2}$, where $\theta \in [0,5\pi ]$ is

• A. $4$
• B. $9$
• C. $3$
• D. $6$

Answer 1

The correct option is C $3$

Given : ${\text{cosec}}^{-1}(\sin \theta +4{\sin }^2\theta )+{\sec }^{-1}(-1+6\sin \theta )=\frac{\pi }{2}$
$\Rightarrow {\sec }^{-1}(-1+6\sin \theta )={\sec }^{-1}(\sin \theta +4{\sin }^2\theta )(\because {\sec }^{-1}x+{\text{cosec}}^{-1}x=\frac{\pi }{2})\Rightarrow \sin \theta +4{\sin }^2\theta =-1+6\sin \theta \Rightarrow 4{\sin }^2\theta -5\sin \theta +1=0\Rightarrow (4\sin \theta -1)(\sin \theta -1)=0\Rightarrow \sin \theta =\frac{1}{4},1$
When $\sin \theta =\frac{1}{4}$
$\sin \theta +4{\sin }^2\theta =-1+6\sin \theta =\frac{1}{2}$
But the domain of ${\sec }^{-1}x$ and ${\text{cosec}}^{-1}x$ is $(-\infty ,-1]\cup [1,\infty )$, so
$\sin \theta =1\Rightarrow \theta =\frac{\pi }{2},\frac{5\pi }{2},\frac{9\pi }{2}(\because x\in [0,5\pi \left]\right)$
Hence, the number of solutions is $3.$