Without using trigonometric tables, prove that:

$\dfrac{\text{cos 80°}}{\text{sin 10°}} + \text{cos 59° cosec 31°} = 2$

To prove,

$\dfrac{\text{cos 80°}}{\text{sin 10°}} + \text{cos 59° cosec 31°} = 2$.

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

$\phantom{\Rightarrow} \dfrac{\text{cos 80°}}{\text{sin 10°}} + \text{cos 59° cosec 31°} \\[1em] \Rightarrow \dfrac{\text{cos 80°}}{\text{sin (90° - 80°)}} + \text{cos 59° cosec (90° - 59°)} \\[1em] \text{As, sin (90 - θ) = cos θ and cosec (90 - θ) = sec θ } \\[1em] \Rightarrow \dfrac{\text{cos 80°}}{\text{cos 80°}} + \text{cos 59° sec 59°} \\[1em] \Rightarrow 1 + \text{cos 59°} \times \dfrac{1}{\text{cos 59°}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.$

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

Hence, proved that $\dfrac{\text{cos 80°}}{\text{sin 10°}} + \text{cos 59° cosec 31°} = 2$.