Without using trigonometric tables, evaluate the following:

$\dfrac{\text{sec 29°}}{\text{cosec 61°}} + \text{2 cot 8° cot 17° cot 45° cot 73° cot 82°} - \text{3 (sin}^2 38° + \text{sin}^2 52°)$.

Solving,

$\Rightarrow \dfrac{\text{sec 29°}}{\text{cosec 61°}} + \text{2 cot 8° cot 17° cot 45° cot 73° cot 82°} - \text{3 (sin}^2 38° + \text{sin}^2 52°) \\[1em] \Rightarrow \dfrac{\text{sec 29°}}{\text{cosec (90° - 29°)}} + \text{2 cot (90° - 82°) cot (90° - 73°) cot 45° cot 73° cot 82°} - \text{3 (sin}^2 38° + \text{sin}^2 (90° - 38°)) \\[1em] \text{As, cosec(90° - θ) = sec θ, cot(90° - θ) = tan θ} \\[1em] \Rightarrow \dfrac{\text{sec 29°}}{\text{sec 29°}} + \text{2 tan 82° tan 73° cot 45°} \times \dfrac{1}{\text{tan 73°}} \times \dfrac{1}{\text{tan 82°}} - \text{3 (sin}^2 38° + \text{cos}^2 38°) \\[1em] \text{As, sin}^2 θ + \text{cos}^2 θ = 1 \\[1em] \Rightarrow 1 + \text{2 cot 45°} - 3 \\[1em] \Rightarrow 1 + 2 - 3 \\[1em] \Rightarrow 0.$

Hence, $\dfrac{\text{sec 29°}}{\text{cosec 61°}} + \text{2 cot 8° cot 17° cot 45° cot 73° cot 82°} - \text{3 (sin}^2 38° + \text{sin}^2 52°)$ = 0.