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)

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