If tanθ + cotθ = 5, then tan²θ + cot²θ is?

Solution (By Examveda Team)

$$\eqalign{ & {\text{Given,}} \cr & {\text{tan}}\theta + \cot \theta = 5 \cr & \Rightarrow {\text{tan}}\theta + \cot \theta = 5 \cr & \Rightarrow {\left( {{\text{tan}}\theta + \cot \theta } \right)^2} = {5^2} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\cot ^2}\theta + 2{\text{tan}}\theta \cot \theta = 25 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = {\text{25}} - {\text{2}} \cr & \left[ {\because {\text{tan}}\theta .\cot \theta = 1} \right] \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = 23 \cr} $$