Without using trigonometric tables, evaluate the following:

$\dfrac{\text{cos 65°}}{\text{sin 25°}} + \dfrac{\text{cos 32°}}{\text{sin 58°}} - \text{sin 28° sec 62° + cosec}^2 30°$

Solving,

$\Rightarrow \dfrac{\text{cos 65°}}{\text{sin (90° - 65°)}} + \dfrac{\text{cos 32°}}{\text{sin (90° - 32°)}} - \text{sin 28°} \times \dfrac{1}{\text{cos 62°}} + \dfrac{1}{\text{sin}^2 30°} \\[1em] \Rightarrow \dfrac{\text{cos 65°}}{\text{cos 65°}} + \dfrac{\text{cos 32°}}{\text{cos 32°}} - \text{sin 28°} \times \dfrac{1}{\text{cos 62°}} + \dfrac{1}{\text{sin}^2 30°} \\[1em] \Rightarrow 1 + 1 - \text{sin 28°} \times \dfrac{1}{cos(90° - 28°)} + \dfrac{1}{\Big(\dfrac{1}{2}\Big)^2} \\[1em] \Rightarrow 2 - \dfrac{\text{sin 28°}}{\text{sin 28°}} + 4 \\[1em] \Rightarrow 2 - 1 + 4 \\[1em] \Rightarrow 5.$

Hence, $\dfrac{\text{cos 65°}}{\text{sin 25°}} + \dfrac{\text{cos 32°}}{\text{sin 58°}} - \text{sin 28° sec 62° + cosec}^2 30°$ = 5.