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.
 A: 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.
A: $$\tan(\pi\cos\theta)=\cot(\pi\sin\theta)\implies \sin(\pi\cos\theta)\sin(\pi\sin\theta)=\cos(\pi\cos\theta)\cos(\pi\sin\theta)\implies$$
$$\cos\left(\pi(\cos\theta-\sin\theta)\right)-\cos\left(\pi(\cos\theta+\sin\theta)\right)=\cos\left(\pi(\cos\theta-\sin\theta)\right)+\cos\left(\pi(\cos\theta+\sin\theta)\right)$$
$$\implies\cos\left(\pi(\cos\theta+\sin\theta)\right)=0\iff\cos\theta+\sin\theta=\frac{2n+1}2\;\;,\;\;n\in\Bbb Z$$
But we can only have $\;n\in\{-2,-1,0,1\}\;$ (why?), so 
$$\sin\theta+\cos\theta=k\iff \sin x\cos\frac\pi4+\sin\frac\pi4\cos\theta=k\cos\frac\pi4\iff$$
$$\iff\sin\left(\theta+\frac\pi4\right)=\frac k{\sqrt2}\;\ldots$$
A: 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,
$\cos\theta + \sin\theta = \pm \frac 1 2\\
\frac 1 {\sqrt 2} \cos\theta + \frac 1 {\sqrt 2} \sin\theta = \pm \frac 1 {2 \sqrt 2}\\
\cos\frac \pi 4 \cos\theta + \sin\frac \pi 4 \sin\theta = \pm \frac 1 {2 \sqrt2}\\
\cos(\theta - \frac \pi 4) = \pm \frac 1 {2 \sqrt 2}$
A: $$\tan (\pi \cos x)=\cot(\pi \sin x)$$
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
