# Tag Info

Actually I just had to solve the following equation to find a solution, consistent with the presence of the "2" in the numerator: $\frac{1}{1+\Lambda (m)}=\frac{2}{1+\sqrt{1+\alpha ^{2}tan_{2}\theta_{m} }}$ $1+\Lambda (m)=\frac{1+\sqrt{1+\alpha ^{2}tan_{2}\theta_{m} }}{2}$ $\Lambda (m)=\frac{-1+\sqrt{1+\alpha ^{2}tan_{2}\theta_{m} }}{2}$