Question

If $\tan \left(\frac{\pi }{2}\sin \theta \right)=\cot \left(\frac{\pi }{2}\cos \theta \right)$, then $\sin (\theta +\frac{\pi }{4})$ can be

• A. $\frac{1}{\sqrt{2}}$
• B. $-1$
• C. $-\frac{1}{\sqrt{2}}$
• D. $1$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{1}{\sqrt{2}}$
C $-\frac{1}{\sqrt{2}}$

$\tan \left(\frac{\pi }{2}\sin \theta \right)=\cot \left(\frac{\pi }{2}\cos \theta \right)$
$\Rightarrow \tan \left(\frac{\pi }{2}\sin \theta \right)=\tan (\frac{\pi }{2}-\frac{\pi }{2}\cos \theta )$
$\Rightarrow \frac{\pi }{2}\sin \theta =n\pi +\frac{\pi }{2}-\frac{\pi }{2}\cos \theta$
$\Rightarrow \frac{\pi }{2}(\sin \theta +\cos \theta )=(2n+1)\frac{\pi }{2},n\in Z$
$\Rightarrow \sin \theta +\cos \theta =2n+1$
$\Rightarrow \frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta =\frac{2n+1}{\sqrt{2}}$
$\Rightarrow \sin (\theta +\frac{\pi }{4})=\frac{2n+1}{\sqrt{2}},n\in Z$
Since $\sin \theta \in [-1,1]$
$\Rightarrow \sin (\theta +\frac{\pi }{4})=\pm \frac{1}{\sqrt{2}}$