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

Answer: (A)

Given, $0\le P,Q\le \frac{\pi }{2}$
and $\sin P+\cos Q=2$ ..... (i)
This equation hold only when, $P=\frac{\pi }{2}$
and $Q=0$
$LHS=\sin P+\cos Q=\sin \frac{\pi }{2}+\cos 0$
$=1+1=2=RHS$
$\therefore \tan \left(\frac{P+Q}{2}\right)=\tan \left(\frac{\frac{\pi }{2}+0}{2}\right)=\tan \frac{\pi }{4}=1$