$$\frac{{{\text{tan}}\theta + \cot \theta }}{{{\text{tan}}\theta - \cot \theta }} = 2,$$   $$\left( {0 \leqslant \theta \leqslant {{90}^ \circ }} \right),$$   then the value of $$\sin \theta $$  is?

Solution (By Examveda Team)

$$\eqalign{ & \frac{{{\text{tan}}\theta + \cot \theta }}{{{\text{tan}}\theta - \cot \theta }} = 2 \cr & {\text{By componendo and dividendo}} \cr & \Rightarrow \frac{{2{\text{tan}}\theta }}{{2{\text{cos}}\theta }} = \frac{3}{1} \cr & \Rightarrow \frac{{\sin \theta }}{{{\text{cos}}\theta }} \times \frac{{\sin \theta }}{{{\text{cos}}\theta }} = 3 \cr & \Rightarrow {\sin ^2}\theta = 3{\text{co}}{{\text{s}}^2}\theta \cr & \Rightarrow {\sin ^2}\theta = 3\left( {1 - {{\sin }^2}\theta } \right) \cr & \Rightarrow 4{\sin ^2}\theta = 3 \cr & \Rightarrow {\sin ^2}\theta \Rightarrow \frac{3}{4} \cr & \Rightarrow {\text{sin }}\theta = \frac{{\sqrt 3 }}{2} \cr & \cr & {\bf{Alternate:}} \cr & \Rightarrow \frac{{{\text{tan}}\theta + \cot \theta }}{{{\text{tan}}\theta - \cot \theta }} = 2 \cr & {\text{By C and D}} \cr & \Rightarrow \frac{{{\text{tan}}\theta }}{{\cot \theta }} = \frac{3}{1} \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = 3 \cr & \Rightarrow {\text{tan}}\theta = \sqrt 3 \cr & \theta = {60^ \circ } \cr & \Rightarrow \sin \theta \cr & \Rightarrow {\text{sin }}{60^ \circ } \cr & \Rightarrow \frac{{\sqrt 3 }}{2} \cr} $$