If $$\frac{{{\text{sec}}\theta + {\text{tan}}\theta }}{{{\text{sec}}\theta - {\text{tan}}\theta }} = 2\frac{{51}}{{79}}{\text{,}}$$     then the value of $$\sin \theta $$  is?

Solution(By Examveda Team)

$$\eqalign{ & {\text{Given, }}\frac{{{\text{sec}}\theta + {\text{tan}}\theta }}{{{\text{sec}}\theta - {\text{tan}}\theta }} = {\text{2}}\frac{{51}}{{79}} \cr & \Rightarrow \frac{{{\text{sec}}\theta + {\text{tan}}\theta }}{{{\text{sec}}\theta - {\text{tan}}\theta }} = \frac{{209}}{{79}} \cr & {\text{By componendo dividendo}} \cr & \left[ {\frac{a}{b} = \frac{c}{d},{\text{ }}\frac{{a + b}}{{a - b}} = \frac{{c + d}}{{c - d}}} \right] \cr & \Rightarrow \frac{{{\text{sec}}\theta + {\text{tan}}\theta + {\text{sec}}\theta - {\text{tan}}\theta }}{{{\text{sec}}\theta + {\text{tan}}\theta - {\text{sec}}\theta + {\text{tan}}\theta }} = \frac{{209 + 79}}{{209 - 79}} \cr & \Rightarrow \frac{{2sec\theta }}{{2{\text{tan}}\theta }} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{{sec\theta }}{{{\text{tan}}\theta }} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{{\frac{1}{{{\text{cos}}\theta }}}}{{\frac{{\sin \theta }}{{{\text{cos}}\theta }}}} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{1}{{\sin \theta }} = \frac{{288}}{{130}} \cr & \Rightarrow {\text{Therefore, }}\sin \theta = \frac{{130}}{{288}} \cr & \Rightarrow \sin \theta = \frac{{65}}{{144}} \cr} $$