Without using trigonometric tables, prove that:

sec 70° sin 20° - cos 20° cosec 70° = 0

To prove,

sec 70° sin 20° - cos 20° cosec 70° = 0

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

$\Rightarrow \text{sec 70° sin 20° - cos 20° cosec 70°} \\[1em] \Rightarrow \dfrac{1}{\text{cos 70°}} \times \text{sin (90° - 70°)} - \text{cos (90° - 70°)} \times \dfrac{1}{\text{sin 70°}} \\[1em] \text{As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ} \\[1em] \Rightarrow \dfrac{1}{\text{cos 70°}} \times \text{cos 70°} - \text{sin 70°} \times \dfrac{1}{\text{sin 70°}} \\[1em] \Rightarrow 1 - 1 \\[1em] \Rightarrow 0.$

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

Hence, proved that sec 70° sin 20° - cos 20° cosec 70° = 0.