<s>[INST] 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.
 [/INST] $$\tan(\pi\cos\theta)=\cot(\pi\sin\theta)\implies \sin(\pi\cos\theta)\sin(\pi\sin\theta)=\cos(\pi\cos\theta)\cos(\pi\sin\theta)\implies$$
$$\cos\left(\pi(\cos\theta-\sin\theta)\right)-\cos\left(\pi(\cos\theta+\sin\theta)\right)=\cos\left(\pi(\cos\theta-\sin\theta)\right)+\cos\left(\pi(\cos\theta+\sin\theta)\right)$$
$$\implies\cos\left(\pi(\cos\theta+\sin\theta)\right)=0\iff\cos\theta+\sin\theta=\frac{2n+1}2\;\;,\;\;n\in\Bbb Z$$
But we can only have $\;n\in\{-2,-1,0,1\}\;$ (why?), so 
$$\sin\theta+\cos\theta=k\iff \sin x\cos\frac\pi4+\sin\frac\pi4\cos\theta=k\cos\frac\pi4\iff$$
$$\iff\sin\left(\theta+\frac\pi4\right)=\frac k{\sqrt2}\;\ldots$$
</s>