Question

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

• 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. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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

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

$\Rightarrow \sin \theta +\cos \theta =2n+1n\in Z$

Also, $\sin \theta +\cos \theta =\sqrt{2}\cos (\theta -\frac{\pi }{4})\Rightarrow -\sqrt{2}\le \sin \theta +\cos \theta \le \sqrt{2}$

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

Hence assertion is correct.

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

Now $-1\le \sin (\theta +\frac{\pi }{4})\le 1\Rightarrow -\sqrt{2}\le \sqrt{2}\sin (\theta +\frac{\pi }{4})\le \sqrt{2}$

$\Rightarrow -\sqrt{2}\le \sin \theta +\cos \theta \le \sqrt{2}$

Hence reason is correct.