Question

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

Answer 1

Given that,

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

If $A$ does not lie in the IInd quadrant then,

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

$\mathrm{cosA}=-cos{60}^0$

$\cos A=\cos ({180}^0+{60}^0)In3rdQuadrant$

$A={240}^0$

If $B$ does not lie in the IIInd quadrant then,

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

$\mathrm{cosB}=-cos{60}^0$

$\mathrm{cosB}=\cos ({180}^0-{60}^0)In3rdQuadrant$

$B={120}^0$

Find the value of

$\frac{4\sin B-3\mathrm{tanA}}{\tan B+\sin A}$

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

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

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

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

$=\frac{4\sqrt{3}-6\sqrt{3}}{2\sqrt{3}+\sqrt{3}}$

$=\frac{-2\sqrt{3}}{3\sqrt{3}}$

$=-\frac{2}{3}$

Hence, this is the answer.