Without using trigonometric tables, evaluate the following:

$\dfrac{\text{sin 25°}}{\text{sec 65°}} + \dfrac{\text{cos 25°}}{\text{cosec 65°}}$.

Solving,

$\Rightarrow \dfrac{\text{sin 25°}}{\text{sec 65°}} + \dfrac{\text{cos 25°}}{\text{cosec 65°}} \\[1em] \Rightarrow \dfrac{\text{sin 25°}}{\dfrac{1}{\text{cos 65°}}} + \dfrac{\text{cos 25°}}{\dfrac{1}{\text{sin 65°}}} \\[1em] \Rightarrow \text{sin 25°.cos 65° + cos 25°.sin 65°} \\[1em] \Rightarrow \text{sin 25°. cos(90° - 25°) + cos 25°. sin(90° - 25°)} \\[1em] \text{As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ} \\[1em] \Rightarrow \text{sin 25°. sin 25° + cos 25°. cos 25°} \\[1em] \Rightarrow \text{sin}^2 25° + \text{cos}^2 25° \\[1em] \Rightarrow 1.$

Hence, $\dfrac{\text{sin 25°}}{\text{sec 65°}} + \dfrac{\text{cos 25°}}{\text{cosec 65°}} = 1$.