If $${\text{A}} \times {\text{tan}}\left( {\theta + {{150}^ \circ }} \right)$$    = $${\text{B}} \times \tan $$ $$\left( {\theta - {{60}^ \circ }} \right){\text{,}}$$   the value of $$\frac{{{\text{A}} - {\text{B}}}}{{{\text{A}} + {\text{B}}}}$$  is?

Solution (By Examveda Team)

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