Prove that :

$\dfrac{\text{sin 60°}}{\text{cos 30°}}$ + $\dfrac{\text{cosec 30°}}{\text{sec 60°}}$ - 4 cos 60° cosec 30° + 2 = 0.

$\text{L.H.S.} = \dfrac{\text{sin 60°}}{\text{cos 30°}} + \dfrac{\text{cosec 30°}}{\text{sec 60°}} - 4 \text{cos 60° cosec 30°} + 2\\[1em] = \dfrac{\text{sin (90° - 30°)}}{\text{cos 30°}} + \dfrac{\text{cosec (90° - 60°)}}{\text{sec 60°}} - 4 \text{cos (90° - 30°) cosec 30°} + 2\\[1em] = \dfrac{\text{cos 30°}}{\text{cos 30°}} + \dfrac{\text{sec 60°}}{\text{sec 60°}} - 4 \text{sin 30° cosec 30°} + 2\\[1em] = \dfrac{\cancel{cos 30°}}{\cancel{cos 30°}} + \dfrac{\cancel{sec 60°}}{\cancel{sec 60°}} - 4 \text{sin 30°} \dfrac{1}{\text{sin 30°}} + 2\\[1em] = 1 + 1 - 4 \cancel{sin 30°} \dfrac{1}{\cancel{sin 30°}} + 2\\[1em] = 2 - 4 + 2\\[1em] = 0$

R.H.S. = 0

∴ L.H.S. = R.H.S.

Hence, $\dfrac{\text{sin 60°}}{\text{cos 30°}}$ + $\dfrac{\text{cosec 30°}}{\text{sec 60°}}$ - 4 cos 60° cosec 30° + 2 = 0.