Question

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

Answer 1

We need to remember the concept of trigonometric transformation formulae and trigonometric ratios and where the signs of cosine and sine will be positive and negative. First find the values of A and B with respect to given conditions and then substitute those values in the required equation.

Complete Step by Step Solution:
The objective of the problem is to find the value of $\dfrac{{4\sin B - 3\tan A}}{{\tan B + \sin A}}$.
Let us consider $\dfrac{{4\sin B - 3\tan A}}{{\tan B + \sin A}}.......\left( 1 \right)$
Given that $\cos A = \cos B = - \dfrac{1}{2}$
Let us take $\cos A = - \dfrac{1}{2}$
We know that the value of $\cos {60^\circ }$ is equal to half.
Now we get $\cos A = - \cos 60$
It is given that A does not lie in the second quadrant so we write $\cos {60^\circ }$ as $\cos \left( {{{180}^\circ } + {{60}^\circ }} \right)$
We get , $\cos A = \cos \left( {{{180}^\circ } + {{60}^\circ }} \right)$
$\Rightarrow \cos A = \cos \left( {{{240}^\circ }} \right)$
Therefore , $A = {240^\circ }$
Now let us take $\cos B = - \dfrac{1}{2}$
We know that the value of $\cos {60^\circ }$ is equal to half.
Now we get $\cos B = - \cos 60$
It is given that B does not lie in the third quadrant so we write $\cos {60^\circ }$ as $\cos \left( {{{180}^\circ } - {{60}^\circ }} \right)$
We get , $\cos B = \cos \left( {{{180}^\circ } - {{60}^\circ }} \right)$
$\Rightarrow \cos B = \cos \left( {{{120}^\circ }} \right)$
Therefore , $B = {120^\circ }$ .
Now substitute the values of A and B in equation 1 we get
$\dfrac{{4\sin B - 3\tan A}}{{\tan B + \sin A}} = \dfrac{{4\sin {{120}^\circ } - 3\tan {{240}^\circ }}}{{\tan {{120}^\circ } + \sin {{240}^\circ }}}$
We write it as
$\Rightarrow \dfrac{{4\sin \left( {{{180}^\circ } - {{60}^\circ }} \right) - 3\tan \left( {{{180}^\circ } + {{60}^\circ }} \right)}}{{\tan \left( {{{180}^\circ } - {{60}^\circ }} \right) + \sin \left( {{{180}^\circ } + {{60}^\circ }} \right)}}$
On solving above we get
$\Rightarrow \dfrac{{4\sin \left( {{{60}^\circ }} \right) - 3\tan \left( {{{60}^\circ }} \right)}}{{\tan \left( {{{60}^\circ }} \right) + \sin \left( {{{60}^\circ }} \right)}}$
Substitute the values of $\sin {60^\circ } = \dfrac{{\sqrt 3 }}{2},\tan {60^\circ } = \sqrt 3$ we get
$\Rightarrow \dfrac{{4 \times \dfrac{{\sqrt 3 }}{2} - 3 \times \sqrt 3 }}{{\sqrt 3 + \dfrac{{\sqrt 3 }}{2}}}$
By taking LCM as two in denominator we get
$\Rightarrow \dfrac{{2\sqrt 3 - 3\sqrt 3 }}{{\dfrac{{2\sqrt 3 + \sqrt 3 }}{2}}}$
On solving above equation we get

\Rightarrow \dfrac{{ - \sqrt 3 }}{{\dfrac{{3\sqrt 3 }}{2}}} \\\ \Rightarrow - \dfrac{{2\sqrt 3 }}{{3\sqrt 3 }} \\\ \Rightarrow - \dfrac{2}{3} \\\ \end{gathered} $$ **Thus the value of $\dfrac{{4\sin B - 3\tan A}}{{\tan B + \sin A}}$ is $$ - \dfrac{2}{3}$$** **Note:** The sine trigonometric function is positive in first and second quadrants in the coordinate axis and in the remaining it is negative only. And for the tan trigonometric function it is positive in the first and third quadrants of the coordinate system. For this type of problem we should remember the trigonometric ratios and formulae. And one should be careful about the algebraic signs and trigonometric signs.