Question

Assertion :If $2\sin \frac{\theta }{2}=\sqrt{1+\sin \theta }+\sqrt{1-\sin \theta }$ then $\frac{\theta }{2}$ lies between $2n\pi +\frac{\pi }{4}$ and $2n\pi +\frac{3\pi }{4}$. Reason: If $\frac{\theta }{2}$ runs from $\frac{\pi }{4}$ to $\frac{3\pi }{4}$, 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:
For $\frac{\pi }{4}\le \frac{\theta }{2}\le \frac{3\pi }{4}$ lies in $1^{st}$ and $2^{nd}$ quadrant and $\sin$ is positive in first two quadrant.
Therefore, $\sin \frac{\theta }{2}>0$

Assertion:$\sqrt{1+\sin \theta }+\sqrt{1-\sin \theta }=\sqrt{{(\sin \frac{\theta }{2}+\cos \frac{\theta }{2})}^2}+\sqrt{{(\sin \frac{\theta }{2}-\cos \frac{\theta }{2})}^2}$

$=\sin \frac{\theta }{2}+\cos \frac{\theta }{2}+\sin \frac{\theta }{2}-\cos \frac{\theta }{2}$ $=2\sin \frac{\theta }{2}$ Only when $\sin \frac{\theta }{2}>\cos \frac{\theta }{2}\Rightarrow 2n\pi +\frac{\pi }{4}\le \frac{\theta }{2}\le 2n\pi +\frac{3\pi }{4}$