Question

If $\mathrm{cosA}=cosB=-\frac{1}{2}$ and A does not lie in the second quadrant and B does not lie in the third quadrant, then find the value of $\frac{4\sin B-3\tan A}{\tan B+\mathrm{sinA}}$

• A. $\frac{1}{5}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $-\frac{1}{3}$

Answer 1

The correct option is D $-\frac{1}{3}$

Consider the given equation.

$\frac{4\sin B-3\tan A}{\tan B+\sin A}$ …... (1)

and $\cos A=\cos B=-\frac{1}{2}$

Now,

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

$\Rightarrow \cos A=-\cos {60}^0$ when $A$ does not lie in the second quadrant.

$\Rightarrow \cos A=\cos ({180}^0+{60}^0)$

$\Rightarrow \cos A=\cos {240}^0$

$\Rightarrow A={240}^0$

Now,

$\mathrm{cosB}=-\frac{1}{2}$

$\Rightarrow \mathrm{cosB}=-\cos {60}^0$ when $B$ does not lie in the third quadrant.

$\Rightarrow \mathrm{cosB}=\cos ({180}^0-{60}^0)$

$\Rightarrow \mathrm{cosB}=\cos {120}^0$

$\Rightarrow B={120}^0$

Substituting the value of $A$ and $B$ in equation (1) and we get,

$\frac{4\sin {120}^0-3\tan {240}^0}{\tan {120}^0+\sin {240}^0}$

$\Rightarrow \frac{4\times \sin ({180}^0-{60}^0)-3\tan ({180}^0+{60}^0)}{\tan ({180}^0-{60}^0)+\sin ({180}^0+{60}^0)}$

$\Rightarrow \frac{4\times \sin {60}^0-3\times \tan {60}^0}{\tan {60}^0+\sin {60}^0}$

$\Rightarrow \frac{4\times \frac{\sqrt{3}}{2}-3\times \sqrt{3}}{\sqrt{3}+\frac{\sqrt{3}}{2}}$

$\Rightarrow \frac{\frac{2\sqrt{3}-3\sqrt{3}}{2}}{\frac{2\sqrt{3}+\sqrt{3}}{2}}$

$\Rightarrow \frac{-\sqrt{3}}{3\sqrt{3}}$

$\Rightarrow \frac{-1}{3}$

Hence, this is the answer.