Question

The general solution of
$tan \left(\frac{\pi}{2} sin \theta\right) = cot \left(\frac{\pi}{2} cos \theta\right)$ is

• A. $\theta = 2r\pi + \frac{\pi}{2}, r \in Z$
• B. $\theta = 2r\pi , r \in Z$
• C. $\theta = 2r\pi +\frac{\pi}{2}$ and $\theta = 2r\pi, r \in Z$
• D. None of the above

Answer 1

Answer: (B)

$\tan \left(\frac{\pi}{2} \sin \theta\right)=\cot \left(\frac{\pi}{2} \cos \theta\right)$
$\Rightarrow \tan \left(\frac{\pi}{2} \sin \theta\right)=\tan \left(\frac{\pi}{2}-\frac{\pi}{2} \cos \theta\right)$
$\Rightarrow \frac{\pi}{2} \sin \theta=r \pi+\frac{\pi}{2}-\frac{\pi}{2} \cos \theta, r \in Z$
$\Rightarrow \sin \theta+\cos \theta=(2 r+1), r \in Z$
$\Rightarrow \frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta=\frac{2 r+1}{\sqrt{2}}, r \in Z$
$\Rightarrow \cos \left(\theta-\frac{\pi}{4}\right)=\frac{2 r+1}{\sqrt{2}}, r \in Z$
$\Rightarrow \cos \left(\theta-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$ or $-\frac{1}{\sqrt{2}}
(for\left.r=0,-1\right)$
$\Rightarrow \theta-\frac{\pi}{4}=2 r \pi \pm \frac{\pi}{4}, r \in Z$
$\Rightarrow \theta=2 r \pi \pm \frac{\pi}{4}+\frac{\pi}{4}, r \in Z$
$\Rightarrow \theta=2 r \pi, 2 r \pi+\frac{\pi}{2}, r \in Z$
But $\theta=2 r \pi+\frac{\pi}{2}, r \in Z$ does not satisfy the given equation.
$\therefore \theta=2 r \pi, r \in Z$