Question

If $\cos A=\frac{4}{5},\cos B=\frac{12}{13},\frac{3\pi }{2}<A,B<2\pi ,$ the values of the following : $\cos (A+B)$ is

• A. $\frac{65}{33}$
• B. $\frac{33}{65}$
• C. $\frac{30}{65}$
• D. $\frac{65}{30}$

Answer 1

Answer: (B)

Since A and B both lie in the IV quadrant, it follows that sin A and sin B are negative.
Therefore,
$\sin A=-\sqrt{1-{\cos }^{2}A}$
$\Rightarrow \sin A=-\sqrt{1-\frac{16}{25}}$
$=-\frac{3}{5}$
and $\sin B=-\sqrt{1-{\cos }^{2}B}$
$\Rightarrow \sin B=-\sqrt{1-\frac{144}{169}}$
$=-\frac{5}{13}$
Now , $\cos \left(A+B\right)$
$=\cos A\cos B-\sin A\sin B$
$=\frac{4}{5}\times \frac{12}{13}-\left(\frac{-3}{5}\right)\left(\frac{-5}{13}\right)$
$=\frac{33}{65}$