If $\text{cot A} = {\sqrt5}$, the value of cosec² A - sec² A is :

1. $\dfrac{5}{24}$
2. $4\dfrac{4}{5}$
3. 5
4. 24

Given:

$\text{cot A} = {\sqrt5}\\[1em] \text{cot A} = \dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{\sqrt5}{1}\\[1em]$

∴ If length of AB = $\sqrt{5}$ x unit, length of BC = x unit.

In Δ ABC,

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

⇒ AC² = $(\sqrt{5}$ x)² + (x)²

⇒ AC² = 5x² + x²

⇒ AC² = 6x²

⇒ AC = $\sqrt{6\text{x}^2}$

⇒ AC = $\sqrt{6}$ x

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

$= \dfrac{AC}{BC} = \dfrac{\sqrt{6}\text{x}}{\text{x}} = \dfrac{\sqrt{6}}{1}$

sec $A = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{\sqrt{6}\text{x}}{\sqrt{5}\text{x}} = \dfrac{\sqrt{6}}{\sqrt{5}}$

Now, cosec² A - sec² A

$= \Big(\dfrac{\sqrt{6}}{1}\Big)^2 - \Big(\dfrac{\sqrt{6}}{\sqrt{5}}\Big)^2\\[1em] = \dfrac{6}{1} - \dfrac{6}{5}\\[1em] = \dfrac{6 \times 5}{1 \times 5} - \dfrac{6 \times 1}{5 \times 1}\\[1em] = \dfrac{30}{5} - \dfrac{6}{5}\\[1em] = \dfrac{30 - 6}{5}\\[1em] = \dfrac{24}{5}\\[1em] = 4\dfrac{4}{5}$

Hence, option 2 is the correct option.