Question

If $\cos A=\frac{5}{13},\tan B=-\frac{15}{8}$, where $A$ is in $IV$ quadrant and $B$ is in $II$ quadrant, then $A+B$ belongs to which quadrant ?

• A. $IV$
• B. $III$
• C. $II$
• D. $I$

Answer 1

The correct option is D $I$

$\cos A=\frac{5}{13}$
As $A$ lies in $IV$ quadrant,
$\sin A=-\frac{12}{13}$
$\tan B=-\frac{15}{8}$
As $B$ lies is in $II$ quadrant,
$\sin B=\frac{15}{17},\cos B=-\frac{8}{17}$
Now,
${270}^{\circ }<A<{360}^{\circ }{90}^{\circ }<B<{180}^{\circ }{360}^{\circ }<A+B<{540}^{\circ }$
So, $A+B$ can lie in $I$ or $II$ quadrant,

$\cos (A+B)=\cos A\cos B-\sin A\sin B\Rightarrow \cos (A+B)=\frac{-40+180}{17\times 13}>0$

Therefore, $A+B$ lies in the $I$ quadrant.