Evaluate :

$\dfrac{\text{cot}^2 41°}{\text{tan}^2 49°} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 15°}$

Solving,

$\Rightarrow \dfrac{\text{cot}^2 41°}{\text{tan}^2 49°} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 15°} \\[1em] \Rightarrow \dfrac{\text{cot}^2 41°}{\text{tan}^2 (90° - 41°)} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 (90° - 75°)}$

By formula,

tan (90° - θ) = cot θ and cos (90° - θ) = sin θ

$\Rightarrow \dfrac{\text{cot}^2 41°}{\text{cot}^2 41°} - 2\dfrac{\text{sin}^2 75°}{\text{sin}^2 75°} \\[1em] \Rightarrow 1 - 2 \\[1em] \Rightarrow -1.$

Hence, $\dfrac{\text{cot}^2 41°}{\text{tan}^2 49°} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 15°}$ = -1