Question

For $0\le P,Q\le \frac{\pi }{2}$, if $\sin P+\cos Q=2$, then the value of $\tan \left(\frac{P+Q}{2}\right)$ is equal to

• A. $1$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{2}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $1$

Given $0\le P,Q\le \frac{\pi }{2}$
and $\sin P+\cos Q=2$
This equation hold only when $P=\frac{\pi }{2}$ and $Q=0$
$2=1+1=\sin \frac{\pi }{2}+\cos 0$

$⟹\sin P+\cos Q=\sin \frac{\pi }{2}+\cos 0$

$⟹P=\frac{\pi }{2},Q=0$
$\therefore$ $\tan \left(\frac{P+Q}{2}\right)=\tan \left(\frac{\frac{\pi }{2}+0}{2}\right)=\tan \frac{\pi }{4}=1$

Hence, $\tan \left(\frac{P+Q}{2}\right)=1$