Question

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

Answer 1

Consider the given data.

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

We know that cosine function is negative in second quadrant and third quadrant, so if A does not lie in 2^nd quadrant, it must i.e. in 3^rd quadrant $ if B does not lie 3^rd quadrant, it must lie in 2^nd quadrant now, we have

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

$\cos A=-\cos \frac{\pi }{3}\Rightarrow \cos (\pi +\frac{\pi }{3})$

$\cos A=\cos \left(\frac{4\pi }{3}\right)$

$A=\frac{4\pi }{3}$

Similarly,

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

$\cos B=-\cos \frac{\pi }{3}\Rightarrow \cos (\pi -\frac{\pi }{3})$

$\cos B=\cos \left(\frac{2\pi }{3}\right)$

$B=\frac{2\pi }{3}$

Therefore,

$\sin A=\sin \frac{4\pi }{3}=-\frac{\sqrt{3}}{2}$

$\tan A=\tan \frac{4\pi }{3}=\sqrt{3}$

$\mathrm{sinB}=\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}$

$\mathrm{tanB}=\tan \frac{2\pi }{3}=-\sqrt{3}$

Since,

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

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

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

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

$=\frac{2}{3}$

Hence, the value is $\frac{2}{3}$.