Question

Find $\theta$:

$\tan \left(\frac{\pi }{2}\sin \theta \right)=\cot \left(\frac{\pi }{2}\cos \theta \right)$.

Answer 1

$\tan \left(\frac{\pi }{2}\sin \theta \right)=\cot \left(\frac{\pi }{2}\cos \theta \right)$
$\therefore \tan \left(\frac{\pi }{2}\sin \theta \right)=\tan (\frac{\pi }{2}-\frac{\pi }{2}\cos \theta )$
$\therefore \frac{\pi }{2}\sin \theta =n\pi +\frac{\pi }{2}-\frac{\pi }{2}\cos \theta$
$\therefore \sin \theta +\cos \theta =2n+1$
Divide by $\sqrt{1+1}$
$\frac{1}{\sqrt{\left(2\right)}}\sin \theta +\frac{1}{\sqrt{\left(2\right)}}\cos \theta =\frac{2n+1}{\sqrt{\left(2\right)}}$
$\cos (\theta -\frac{\pi }{4})=\frac{2n+1}{\sqrt{\left(2\right)}}=\cos \alpha$, say
$\theta -\frac{\pi }{4}=2r\pi \pm \alpha$
$\therefore \theta =2r\pi +\frac{\pi }{4}\pm {\cos }^{-1}\frac{2n+1}{\sqrt{\left(2\right)}}$
where $n=0$ or $-1$ only as $\cos \alpha \le 1;r\in I$
$\therefore \theta =2r\pi +\pi /4\pm \pi /4$
or $2r\pi +\pi /4\pm 3\pi /4$
$\because {\cos }^{-1}(-\frac{1}{\sqrt{2}})=\frac{3\pi }{4},r\in I$.
Hence the only solution is $\theta =2r\pi$ or $2r\pi +\pi$. We reject other values as they will lead to $\infty =\infty$.