Question

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

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

Answer 1

Answer: (B)

$\theta =2r\pi ,rϵZ$

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