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.
$$\tan (\pi \cos x)=\cot(\pi \sin x)$$

or,

$$\tan (\pi \cos x)=\tan (\frac{\pi}{2} - \pi \sin x)$$ [∵, tan(π/2+Ф)=cotФ]

or, πcosx=π/2-πsinx

or, π(sinx+cosx)=π/2

or, sinx+cosx=1/2

or, (sinx+cosx)1/√2=1/2√2

or, cosx(1/√2)+sinx(1/√2)=1/2√2

or, cosxcosπ/4+sinxsinπ/4=1/2√2

or, cos(x-π/4)=1/2√2 

∴, a)1/2√2 is the right answer