Question

If $\frac{\cos A}{3}=\frac{\cos B}{4}=\frac{1}{5},-\frac{\pi }{2}<A<0,-\frac{\pi }{2}<B<0$ then value of $2\sin A+4\sin B$ is

• A. $4$
• B. $-2$
• C. $-4$
• D. $0$

Answer 1

Answer: (C)

$-4$

$cosA=\frac{3}{5}\Rightarrow \sin A=-\frac{4}{5}\left\{\begin{array}{c}\because \sin A=\sqrt{1-{\cos }^{2}A}\end{array}\right\}$
and $cosB=\frac{4}{5}\Rightarrow \sin B=-\frac{3}{5}$

Therefore,$2\sin A+4\sin B$

$=2\left(\begin{array}{c}-\frac{4}{5}\end{array}\right)+4\left(\begin{array}{c}-\frac{3}{5}\end{array}\right)=-\frac{20}{5}=-4$