Without using trigonometric tables, prove that:

$\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - 8\text{ sin}^2 30° = 0$

To prove,

$\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - 8\text{ sin}^2 30° = 0$

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

$\Rightarrow \dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - 8\text{ sin}^2 30° \\[1em] \Rightarrow \dfrac{\text{cos (90° - 20°)}}{\text{sin 20°}} + \dfrac{\text{cos (90° - 31°)}}{\text{sin 31°}} - 8\text{ sin}^2 30° \\[1em] \text{As, cos (90 - θ) = sin θ} \\[1em] \Rightarrow \dfrac{\text{sin 20°}}{\text{sin 20°}} + \dfrac{\text{sin 31°}}{\text{sin 31°}} - 8 \times \Big(\dfrac{1}{2}\Big)^2 \\[1em] \Rightarrow 1 + 1 - 2 \\[1em] \Rightarrow 0.$

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

Hence, proved that $\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - 8\text{ sin}^2 30° = 0$.