Without using trigonometric tables, evaluate the following:

$\Big(\dfrac{\text{tan 25°}}{\text{cosec 65°}}\Big)^2 + \Big(\dfrac{\text{cot 25°}}{\text{sec 65°}}\Big)^2 + \text{2 tan 18° tan 45° tan 72°}.$

We need to find the value of,

$\Big(\dfrac{\text{tan 25°}}{\text{cosec 65°}}\Big)^2 + \Big(\dfrac{\text{cot 25°}}{\text{sec 65°}}\Big)^2 +$ 2 tan 18° tan 45° tan 72°.

The above equation can be written as,

$\Rightarrow \Big[\dfrac{\text{tan 25°}}{\text{cosec (90 - 25)°}}\Big]^2 + \Big[\dfrac{\text{cot 25°}}{\text{sec (90 - 25)°}}\Big]^2 + 2 \text{ tan } 18° \times 1 \times \text{tan } (90 - 18)° \\[1em]$

As, sec(90° - θ) = cosec θ, tan(90° - θ) = cot θ and cosec(90° - θ) = sec θ. Using in above equation we get,

$\Rightarrow \Big(\dfrac{\text{tan 25°}}{\text{sec 25°}}\Big)^2 + \Big(\dfrac{\text{cot 25°}}{\text{cosec 25°}}\Big)^2 + 2 \text{ tan } 18° \times 1 \times \text{cot } 18° \\[1em] \Rightarrow \Bigg(\dfrac{\dfrac{\text{sin 25°}}{\text{cos 25°}}}{\dfrac{1}{\text{cos 25°}}}\Bigg)^2 + \Bigg(\dfrac{\dfrac{\text{cos 25°}}{\text{sin 25°}}}{\dfrac{1}{\text{sin 25°}}}\Bigg)^2 + 2\text{ tan } 18° \times \dfrac{1}{\text{tan 18°}} \\[1em] \Rightarrow \text{sin}^2 25° + \text{cos}^2 25° + 2 \\[1em] \Rightarrow 1 + 2 = 3.$

Hence, the value of the equation is 3.