Without using trigonometric tables, prove that:

$\dfrac{\text{sin 50°}}{\text{cos 40°}} + \dfrac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0$.

To prove,

$\dfrac{\text{sin 50°}}{\text{cos 40°}} + \dfrac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0$.

Solving, L.H.S. of the equation we get :

$\phantom{\Rightarrow} \dfrac{\text{sin 50°}}{\text{cos 40°}} + \dfrac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0. \\[1em] \Rightarrow \dfrac{\text{sin (90° - 40°)}}{\text{cos 40°}} + \dfrac{\text{cosec (90° - 50°)}}{\text{sec 50°}} - \text{4 cos 50° cosec (90° - 50°) + 2} = 0 \\[1em]$

We know that,

sin (90 - θ) = cos θ

and

cosec (90 - θ) = sec θ

$\Rightarrow \dfrac{\text{cos 40°}}{\text{cos 40°}} + \dfrac{\text{sec 50°}}{\text{sec 50°}} - \text{4 cos 50° sec 50°} + 2 \\[1em] \Rightarrow 1 + 1 - 4 \times \text{cos 50°} \times \dfrac{1}{\text{cos 50°}} + 2 \\[1em] \Rightarrow 2 - 4 + 2 \\[1em] \Rightarrow 0.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin 50°}}{\text{cos 40°}} + \dfrac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0$.