Question

If $\tan A=-\frac{1}{2}$ and $\tan B=-\frac{1}{3}$. (where $A,B>0$), then $A+B$ can be

• A. $\frac{\pi }{4}$
• B. $\frac{3\pi }{4}$
• C. $\frac{5\pi }{4}$
• D. $\frac{7\pi }{4}$

Answer 1

The correct option is B $\frac{3\pi }{4}$

We have,

$\tan A=-\frac{1}{2}$

$\tan B=-\frac{1}{3}$

Since,

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

Thus,

$\tan (A+B)=\frac{-\frac{1}{2}-\frac{1}{3}}{1-(-\frac{1}{2})(-\frac{1}{3})}$

$\tan (A+B)=\frac{-\left(\frac{5}{6}\right)}{1-\left(\frac{1}{6}\right)}$

$\tan (A+B)=\frac{-\left(\frac{5}{6}\right)}{\left(\frac{5}{6}\right)}$

$\tan (A+B)=-1$

$\tan (A+B)=\tan \left(\frac{3\pi }{4}\right)$

$A+B=\frac{3\pi }{4}$

Hence, this is the answer.