If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the value of $\cos(\theta -\frac{\pi}{4})$? Problem : 

If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the
  value of $\cos(\theta -\frac{\pi}{4})$?

My approach : 
Solution: $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$ 
$\Rightarrow \tan(\pi \cos\theta) = \tan \{ \frac{\pi}{2} - (\pi \sin\theta) \} $
$\Rightarrow \pi \cos\theta = \frac{\pi}{2} - (\pi \sin\theta)$
$\Rightarrow \frac{1}{2} =\frac{1}{\sqrt{2}}[\sin\frac{\pi}{4} \cos\theta + \cos\frac{\pi}{4} \sin\theta] $
$\Rightarrow \frac{1}{\sqrt{2}} = \sin(\frac{\pi}{4} + \theta)$
$\Rightarrow \frac{\pi}{4} = \frac{\pi}{4} + \theta$
$\Rightarrow \theta = 0$ 
$\therefore \cos(\theta - \frac{\pi}{4})$ = $\frac{1}{\sqrt{2}}$ But this is wrong answer.. please suggest where I am wrong... thanks.

I would use
$$\frac{\sin{(\pi \cos{\theta})}}{\cos{(\pi \cos{\theta})}} = \frac{\cos{(\pi \sin{\theta})}}{\sin{(\pi \sin{\theta})}}$$
from which I get
$$\cos{[\pi (\cos{\theta}+\sin{\theta})]}=0$$
or, in one case,
$$\pi (\cos{\theta}+\sin{\theta}) = \frac{\pi}{2}$$
or,
$$\sqrt{2} \pi \cos{\left ( \theta-\frac{\pi}{4}\right )} = \frac{\pi}{2}$$
You can take it from here.
