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.

It is given that,
$\tan(\pi \cos\theta) =\cot(\pi \sin\theta)\\
\frac {\sin(\pi \cos\theta)} {\cos(\pi \cos\theta)} = \frac {\cos(\pi \sin\theta)} {\sin(\pi \sin\theta)}\\
\sin(\pi \cos\theta)\sin(\pi \sin\theta) - \cos(\pi \sin\theta)\cos(\pi \cos\theta) = 0\\
\cos(\pi \cos\theta + \pi \sin\theta) = \cos(\frac \pi 2)$
Therefore,
$\pi \cos\theta + \pi \sin\theta = 2n\pi \pm \frac \pi 2 , [n \in Z]\\
\cos\theta + \sin\theta = 2n \pm \frac 1 2$
Since $\sin\theta + \cos\theta$ always lies between $\sqrt 2$ and $-\sqrt 2$, $n$ can only assume the value $0$. Thus,
$\cos\theta + \sin\theta = \pm \frac 1 2\\
\frac 1 {\sqrt 2} \cos\theta + \frac 1 {\sqrt 2} \sin\theta = \pm \frac 1 {2 \sqrt 2}\\
\cos\frac \pi 4 \cos\theta + \sin\frac \pi 4 \sin\theta = \pm \frac 1 {2 \sqrt2}\\
\cos(\theta - \frac \pi 4) = \pm \frac 1 {2 \sqrt 2}$
