If $${\text{co}}{{\text{s}}^4}\theta - {\sin ^4}\theta = \frac{2}{3}{\text{,}}$$     then the value of $${\text{2co}}{{\text{s}}^2}\theta - 1$$   is?

Solution(By Examveda Team)

$$\eqalign{ & {\text{co}}{{\text{s}}^4}\theta - {\sin ^4}\theta = \frac{2}{3} \cr & \left[ {{a^4} - {b^4} = \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right)} \right] \cr & \Rightarrow \left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right)\left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right) = \frac{2}{3} \cr & \left[ {{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right] \cr & \Rightarrow 1 \times \left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right) = \frac{2}{3} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - \left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) = \frac{2}{3} \cr & \left[ {{{\sin }^2}\theta = 1 - {\text{co}}{{\text{s}}^2}\theta } \right] \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta - 1 = \frac{2}{3} \cr} $$