If 5 cot θ = 12, find the value of :

cosec θ + sec θ

Given:

5 cot θ = 12

⇒ $\text{cot θ} = \dfrac{12}{5}$

$⇒ \text{cot θ} = \dfrac{Base}{Perpendicular} = \dfrac{12}{5}\\[1em]$

∴ If length of BC = 5x unit, length of AB = 12x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169x²

⇒ AC = $\sqrt{169 \text{x}^2}$

⇒ AC = 13x

cosec θ = $\dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AC}{CB} = \dfrac{13x}{5x} = \dfrac{13}{5}$

sec θ = $\dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{13x}{12x} = \dfrac{13}{12}$

cosec θ + sec θ

$= \dfrac{13}{5} + \dfrac{13}{12}\\[1em] = \dfrac{13 \times 12}{5 \times 12} + \dfrac{13 \times 5}{12 \times 5}\\[1em] = \dfrac{156}{60} + \dfrac{65}{60}\\[1em] = \dfrac{156 + 65}{60}\\[1em] = \dfrac{221}{60}\\[1em] = 3\dfrac{41}{60}$

Hence, cosec θ + sec θ = $3\dfrac{41}{60}$.