Question

If $\cos A = - \frac{60}{61} $ and $ \tan B =- \frac{7}{24} $ and neither $A$ nor $B$ is in the second quadrant, then the angle $ A + \frac{B}{2} $ lies in the quadrant .

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

It is given that $\cos A=-\frac{60}{61}$ and $\tan B=\frac{-7}{24}$ and neither $A$ nor $B$ is in the second quadrant, then $A$ and $B$ will be in third and fourth quadrant respectively.
So, $\frac{B}{2}$ will be in second quadrant.
Now, $\tan B=-\frac{7}{24}$
$\Rightarrow \frac{2 \tan \frac{B}{2}}{1-\tan ^{2} \frac{B}{2}}=-\frac{7}{24}$
$\Rightarrow 7 \tan ^{2} \frac{B}{2}-48 \tan \frac{B}{2}-7=0$
$\Rightarrow \left(7 \tan \frac{B}{2}+1\right)\left(\tan \frac{B}{2}-7\right)=0$
$\Rightarrow \tan \frac{B}{2}=-\frac{1}{7}$
$\left[\because \frac{B}{2} \in \text { IInd quadrant }\right]$
$\because \tan \left(A+\frac{B}{2}\right)=\frac{\tan A+\tan \frac{B}{2}}{1-\tan A \tan \frac{B}{2}}$
$=\frac{\frac{11}{60}-\frac{1}{7}}{1+\frac{11}{420}}$
[$\because \cos A=-\frac{60}{6]}$ and $A \in IIIrd$ Quadrant $\therefore \tan A=\frac{11}{60}$]
$=\frac{77-60}{420+11}=\frac{17}{431}$
$\because A \in$ IIIrd quadrant and $\frac{B}{2} \in\left(\frac{3 \pi}{4}, \pi\right)$
and $\tan \left(A+\frac{B}{2}\right)$ is positive.
$\therefore \left(A+\frac{B}{2}\right) \in$ Ist quadrant.