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If $\tan (\pi \cos \theta)=\cot (\pi \sin \theta)$, then a value of $\cos \left(\theta-\frac{\pi}{4}\right)$ among the following is
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Verified Answer
The correct answer is:
$\frac{1}{2 \sqrt{2}}$
Given, $\tan (\pi \cos \theta)=\cot (\pi \sin \theta)$
$$
\begin{aligned}
& \Rightarrow \quad \tan (\pi \cos \theta)=\tan \left\{\frac{\pi}{2}-\pi \sin \theta\right\} \\
& \Rightarrow \quad \pi \cos \theta=\frac{\pi}{2}-\pi \sin \theta \\
& \Rightarrow \quad \sin \theta+\cos \theta=\frac{1}{2} \\
& \Rightarrow \quad \frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta=\frac{1}{2 \sqrt{2}} \\
& \Rightarrow \quad \cos \theta \cdot \cos \frac{\pi}{4}+\sin \theta \cdot \sin \frac{\pi}{4}=\frac{1}{2 \sqrt{2}} \\
& \Rightarrow \quad \cos \left(\theta-\frac{\pi}{4}\right)=\frac{1}{2 \sqrt{2}}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad \tan (\pi \cos \theta)=\tan \left\{\frac{\pi}{2}-\pi \sin \theta\right\} \\
& \Rightarrow \quad \pi \cos \theta=\frac{\pi}{2}-\pi \sin \theta \\
& \Rightarrow \quad \sin \theta+\cos \theta=\frac{1}{2} \\
& \Rightarrow \quad \frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta=\frac{1}{2 \sqrt{2}} \\
& \Rightarrow \quad \cos \theta \cdot \cos \frac{\pi}{4}+\sin \theta \cdot \sin \frac{\pi}{4}=\frac{1}{2 \sqrt{2}} \\
& \Rightarrow \quad \cos \left(\theta-\frac{\pi}{4}\right)=\frac{1}{2 \sqrt{2}}
\end{aligned}
$$
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