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. 
 A: Rearrange into sines and cosines. Form a cosine addition identity.
\begin{align}
\tan(\pi\cos t)\tan(\pi\sin t) &= 1
\\
\sin(\pi\cos t)\sin(\pi\sin t) &= \cos(\pi\cos t)\cos(\pi\sin t)
\\
0 &= \cos(\pi\cos t)\cos(\pi\sin t) - \sin(\pi\cos t)\sin(\pi\sin t)
\\
0 &= \cos(\pi\cos t + \pi\sin t)
\\
0 &= \cos(\pi[\cos t + \sin t])
\end{align}
The argument of the cosine in the previous line must be an odd integer multiple of $\pi/2$, so we get
\begin{align}
\cos t + \sin t &= \frac{2n+1}{2} 
\end{align}
where $n\in\mathbb{Z}$ (EDIT: More accurately, $n \in \{-1,0\}$ to be in the range of $\cos t + \sin t$). Use the cosine subtraction formula to achieve the desired result.
\begin{align}
\frac{1}{\sqrt2}\cos t + \frac{1}{\sqrt2}\sin t &= \frac{2n+1}{2\sqrt{2}}
\\
\cos (t)\cos(\pi/4) + \sin (t)\sin(\pi/4) &= \frac{2n+1}{2\sqrt{2}}
\\
\\
\cos(t -\pi/4) &= \frac{2n+1}{2\sqrt{2}}
\end{align}
EDIT: Following lab bhattacharjee's comment, the range of cosine restricts the possible solutions to 
$$\cos(t -\pi/4) = \frac{\pm1}{2\sqrt{2}}.$$
It doesn't appear like the answer is unique to me. 
A: 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 $$\sin\theta + \cos\theta =\frac{1}{2},$$
$$\cos\left(\theta-\frac{\pi}{4}\right)=\frac{1}{2\sqrt{2}}.$$
A: Hint -
$\tan (\pi \cos \theta) =\cot (\pi \sin \theta)$
$\tan(\pi \cos \theta) =\tan (\frac{\pi}2 - \pi \sin \theta)$
$\pi \cos \theta =\frac{\pi}2 - \pi \sin \theta$
$\cos \theta = \frac 12 - \sin \theta$
$\sin \theta + \cos \theta = \frac 12$
Now we have,
$\cos\left(\theta-\frac{\pi}{4}\right)$
$= \cos\theta \cos \frac{\pi}4 +\sin\theta \sin \frac{\pi}4$
$= \cos\theta \frac 1{\sqrt{2}} + \sin\theta \frac 1{\sqrt{2}}$
$= \frac1{\sqrt 2} \left( \cos\theta +\sin\theta \right)$
$ = \frac 1{\sqrt2} \left(\frac 12\right)$
$= \frac 1{2\sqrt2}$
A: $$\begin{align}
\tan (\pi \cos \theta) &= \cot (\pi \sin \theta) \\
\implies \frac{\sin(\pi \cos \theta)}{\cos(\pi \cos \theta)} &= \frac{\cos(\pi \sin \theta)}{\sin(\pi \sin \theta)} \\
\implies \sin(\pi \cos \theta)\sin(\pi \sin \theta) &= \cos(\pi \sin \theta)\cos(\pi \cos \theta) \\
\end{align}$$
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, $\cos(\theta - \pi/4) = \sin(\theta + \pi/4) = \frac{1}{2\sqrt{2}}$.
