If θ be an acute angle and 7sin²θ + 3cos²θ = 4, then the value of tanθ is?

Solution (By Examveda Team)

$$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\cos ^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {1 - {{\sin }^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} \cr & \Rightarrow \sin \theta = \sin {30^ \circ } \cr & \Rightarrow \theta = {30^ \circ } \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put}}\theta = {30^ \circ } \cr & {\text{7}} \times \sin^2 {30^ \circ } + 3{\cos ^2}{30^ \circ } = 4 \cr & \Rightarrow 7 \times \frac{1}{4} + 3 \times \frac{3}{4} = 4 \cr & \Rightarrow \frac{7}{4} + \frac{9}{4} = 4 \cr & \Rightarrow \frac{{16}}{4} = 4 \cr & \Rightarrow 4 = 4\left( {{\text{Satisfied}}} \right) \cr & \therefore \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr} $$