If tan²α = 1 + 2tan²β (α, β are positive acute angles), then √2cosα - cosβ is equal to?

Solution (By Examveda Team)

$$\eqalign{ & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr & \Rightarrow {\text{se}}{{\text{c}}^2}\alpha - 1 = 1 + 2\left( {{\text{se}}{{\text{c}}^2}\beta - 1} \right) \cr & \Rightarrow {\sec ^2}\alpha - 1 = 2{\sec ^2}\beta - 1 \cr & \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}\alpha }} = \frac{2}{{{\text{co}}{{\text{s}}^2}\beta }} \cr & \Rightarrow \sqrt 2 {\text{cos}}\alpha = {\text{cos}}\beta \cr & \therefore \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta = 0 \cr & \cr & {\bf{Alternate:}} \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr & {\text{Put }}\beta = {45^ \circ } \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2.{\text{ta}}{{\text{n}}^2}{45^ \circ } \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 3 \cr & {\text{ tan }}\alpha = \sqrt 3 \cr & \alpha = {60^ \circ } \cr & {\text{Put }}\alpha = {60^ \circ },{\text{and }}\beta = {45^ \circ } \cr & = \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta \cr & = \sqrt 2 {\text{cos }}{60^ \circ } - {\text{cos }}{45^ \circ } \cr & = \sqrt 2 \times \frac{1}{2} - \frac{1}{{\sqrt 2 }} \cr & = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr & = 0 \cr} $$