If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos\left (\theta -\frac{\pi}{4}\right)$ If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos \left(\theta -\frac{\pi}{4}\right).$
I could not get any idea to solve.  However I tried by using $\theta =0^\circ $. But could not get the answer. 

$$\begin{align}
\tan (\pi \cos \theta) &= \cot (\pi \sin \theta) \\
\implies \frac{\sin(\pi \cos \theta)}{\cos(\pi \cos \theta)} &= \frac{\cos(\pi \sin \theta)}{\sin(\pi \sin \theta)} \\
\implies \sin(\pi \cos \theta)\sin(\pi \sin \theta) &= \cos(\pi \sin \theta)\cos(\pi \cos \theta) \\
\end{align}$$
and then
$$\begin{align}
0 &= \cos(\pi \sin \theta)\cos(\pi \cos \theta) - \sin(\pi \cos \theta)\sin(\pi \sin \theta) \\
&= \cos(\pi\sin\theta + \pi\cos\theta) \\
&= \cos(\pi(\sin\theta + \cos\theta)) \\
\implies \frac{\pi}{2} &= \pi(\sin\theta + \cos\theta) \\
\implies \frac{1}{2} &= \sin\theta + \cos\theta \\
\end{align}$$
Now we use the fact that $\sin \pi/4 = \cos \pi/4 = \frac{1}{\sqrt{2}}$.
$$\begin{align}
\sin\theta + \cos\theta &= \sqrt{2}\bigg(\frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta\bigg) \\
&= \sqrt{2}(\sin\theta\cos \pi/4 + \cos\theta\sin \pi/4) \\
&= \sqrt{2}\sin(\theta + \pi/4) \\
&= \frac{1}{2}
\end{align}$$
Finally, $\cos(\theta - \pi/4) = \sin(\theta + \pi/4) = \frac{1}{2\sqrt{2}}$.
