If θ is a positive acute angle and 4cos²θ - 1 = 0, then the value of tan(θ - 15°) is equal to?

$$\eqalign{ & {\text{4co}}{{\text{s}}^2}\theta - 1 = 0 \cr & \Rightarrow {\text{4co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr & \Rightarrow \cos \theta = \frac{1}{2} = \cos {60^ \circ } \cr & \theta = {60^ \circ } \cr & \Rightarrow \tan \left( {\theta - {{15}^ \circ }} \right) \cr & \Rightarrow \tan \left( {{{60}^ \circ } - {{15}^ \circ }} \right) \cr & \Rightarrow \tan {45^ \circ } \cr & \Rightarrow 1 \cr} $$