Without using trigonometric tables, prove that:

sec 31° sin 59° + cos 31° cosec 59° = 2.

To prove,

sec 31° sin 59° + cos 31° cosec 59° = 2

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

⇒ sec 31° sin 59° + cos 31° cosec 59

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

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

Hence, proved that sec 31° sin 59° + cos 31° cosec 59° = 2.