Question

Assertion :(A) : If $tan\left(\begin{array}{c}\frac{\pi }{2}sin\theta \end{array}\right)=cot\left(\begin{array}{c}\frac{\pi }{2}cos\theta \end{array}\right)$ then $sin\theta +cos\theta =\pm \sqrt{2}$ Reason: (R) :$-\sqrt{2}\le sin\theta +cos\theta \le \sqrt{2}$

• A. Both (A) and (R) are individually true and (R) is the correct explanation of (A)
• B. (R) is not the correct explanation of (A)
• C. (A) is true but (R) is false
• D. (A) is false but (R) is true

Answer 1

The correct option is D (A) is false but (R) is true

$\tan \left(\frac{\pi }{2}\sin \theta \right)=\cot \left(\frac{\pi }{2}cos\theta \right)$
$\Rightarrow \tan \left(\frac{\pi }{2}\sin \theta \right)=\tan (\frac{\pi }{2}-\frac{\pi }{2}cos\theta )$
$\Rightarrow \frac{\pi }{2}\sin \theta =\frac{\pi }{2}-\frac{\pi }{2}cos\theta$
$\Rightarrow \frac{\pi }{2}\sin \theta +\frac{\pi }{2}cos\theta =\frac{\pi }{2}$
$\Rightarrow \sin \theta +cos\theta =1$
As $\Rightarrow -\sqrt{a^2{b}^2}\le a\sin A+bcosA\le \sqrt{a^2{b}^2}$
$\Rightarrow -\sqrt{2}\le \sin \theta +cos\theta \le \sqrt{2}$

Therefore assertion is false and reason is true.