Question

Assertion :If $\tan \left(\frac{\pi }{2}\sin \theta \right)=\cot \left(\frac{\pi }{2}\cos \theta \right)$, then $\sin (\theta +\frac{\pi }{4})=\pm 1$ Reason: $-\sqrt{2}\le \sin (\theta +\frac{\pi }{4})\le \sqrt{2}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is false but Reason is true

Answer 1

The correct option is D Assertion is false but Reason is true

$\tan \left(\frac{\pi }{2}\sin \theta \right)=\cot \left(\frac{\pi }{2}\cos \theta \right)$

$=\tan (\frac{\pi }{2}-\frac{\pi }{2}\cos \theta )$
Hence

$\frac{\pi }{2}\sin \theta =\frac{\pi }{2}-\frac{\pi }{2}\cos \theta$

$\sin \theta +\cos \theta =1$

$\sqrt{2}(\frac{1}{\sqrt{2}}\sin \theta +\cos \theta \frac{1}{\sqrt{2}})=1$

$\sin (\theta +\frac{\pi }{4})=\frac{1}{\sqrt{2}}$

Hence assertion is false.

AS $-2>\sin \theta +\cos \theta <2$

$-\frac{2}{\sqrt{2}}\le \frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta <\frac{2}{\sqrt{2}}$

$-\sqrt{2}<\sin (\theta +\frac{\pi }{4})<\sqrt{2}$