If A × tan ( + 150^ ) = B × ( - 60^ ) , the value of A - B A + B is?

Answer: Option A

$$\begin{array}{rl}& \text{A}\times \text{tan}\left(\theta +{150}^{\circ }\right)=\text{B}\times \tan \left(\theta -{60}^{\circ }\right)\\ & \frac{\text{A}}{\text{B}}=\frac{\tan \left(\theta -{60}^{\circ }\right)}{\tan \left(\theta +{150}^{\circ }\right)}\\ & \text{Put }\theta =90^{\circ }\\ & \frac{\text{A}}{\text{B}}=\frac{\tan \left({90}^{\circ }-{60}^{\circ }\right)}{\tan \left({90}^{\circ }+{150}^{\circ }\right)}\\ & \frac{\text{A}}{\text{B}}=\frac{\tan {30}^{\circ }}{\tan \left({180}^{\circ }+{60}^{\circ }\right)}\\ & \frac{\text{A}}{\text{B}}=\frac{\tan {30}^{\circ }}{\tan {60}^{\circ }}\\ & \frac{\text{A}}{\text{B}}=\frac{1}{3}\\ & \text{then, }\frac{\text{A}+\text{B}}{\text{A}-\text{B}}=-\frac{4}{2}\\ & \Rightarrow \frac{\text{A}+\text{B}}{\text{A}-\text{B}}=-2\\ & \Rightarrow \frac{\text{A}-\text{B}}{\text{A}+\text{B}}=-\frac{1}{2}\\ & \text{Put in option (i)}\\ & -\frac{\sin {90}^{\circ }}{2}=-\frac{1}{2}\\ & \text{So, option (A) is correct }\end{array}$$