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. 

Since $$\tan(\pi\cos\theta)=\cot(\pi\sin\theta),$$
we can write $$\tan(\pi\cos\theta)=\cot(90-(\pi\cos\theta)).$$
Thus  $$\cot(90-(\pi\cos\theta))=\cot(\pi\sin\theta)$$
$$\therefore\quad 90-(\pi\cos\theta)=\pi\sin\theta$$
$$90=\pi(\sin\theta+\cos\theta)$$
$$\frac{90}{\pi}=\sin\theta+\cos\theta$$
$$\frac{1}{2}=\sin\theta +\cos\theta$$
Now
\begin{align*}
\cos\left(\theta-\frac{\pi}{4}\right)&=\cos\theta \cos45^{\circ}+\sin\theta \sin45^{\circ}\\
&=\frac{\cos\theta}{\sqrt{2}}+\frac{\sin\theta}{\sqrt{2}}\\
&=\frac{\cos\theta +\sin\theta}{\sqrt{2}}
\end{align*}
Since $$\sin\theta + \cos\theta =\frac{1}{2},$$
$$\cos\left(\theta-\frac{\pi}{4}\right)=\frac{1}{2\sqrt{2}}.$$
