Question

Assertion :If $2\sin \left(\frac{\theta }{2}\right)=\sqrt{1+\sin \theta }+\sqrt{1-\sin \theta }$ then $\frac{\theta }{2}$ lies between $2n\pi +\frac{\pi }{4}$ and $2n\pi +\frac{3\pi }{4}$. Reason: If $\frac{\pi }{4}\le \theta \le \frac{3\pi }{4}$ then $\sin \frac{\theta }{2}>0$.

• 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 incorrect but Reason is correct

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Reason:
$\sin x>0$ for $0<x<\pi$
Hence, $\sin \frac{\theta }{2}>0$ for $0<\frac{\theta }{2}<\pi \Rightarrow 0<\theta <2\pi$

Assertion:
$2\sin \left(\frac{\theta }{2}\right)=\sqrt{{(\cos \left(\frac{\theta }{2}\right)+\sin \left(\frac{\theta }{2}\right))}^2}+\sqrt{{(\cos \left(\frac{\theta }{2}\right)+\sin \left(\frac{\theta }{2}\right))}^2}=|\cos \left(\frac{\theta }{2}\right)+\sin \left(\frac{\theta }{2}\right)|+|\cos \left(\frac{\theta }{2}\right)-\sin \left(\frac{\theta }{2}\right)|$
$\Rightarrow \cos \left(\frac{\theta }{2}\right)+\sin \left(\frac{\theta }{2}\right)>0$ and $\cos \left(\frac{\theta }{2}\right)-\sin \left(\frac{\theta }{2}\right)<0$
$\Rightarrow \sin (\frac{\theta }{2}+\frac{\pi }{4})>0$ and $\cos (\frac{\theta }{2}+\frac{\pi }{4})<0$
$\Rightarrow 2n\pi +\frac{\pi }{2}<\frac{\theta }{2}+\frac{\pi }{4}<2n\pi +\pi \Rightarrow 2n\pi +\frac{\pi }{4}<\frac{\theta }{2}<2n\pi +\frac{3\pi }{4}$

Therefore assertion and reason both are true but reason is not correct expalintion of assertion