If cot $A = \dfrac{5}{12}$, the value of cot² A - cosec² A is :

1. 1

2. 2

3. -2

4. -1

Given:

cot $A = \dfrac{5}{12}$

i.e. $\dfrac{Base}{Perpendicular} = \dfrac{5}{12}$

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

In Δ ABC,

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

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169²

⇒ AC = $\sqrt{169\text{x}^2}$

⇒ AC = 13x

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

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

Now, cot² A - cosec² A

$= \Big(\dfrac{5}{12}\Big)^2 - \Big(\dfrac{13}{12}\Big)^2\\[1em] = \Big(\dfrac{25}{144}\Big) - \Big(\dfrac{169}{144}\Big)\\[1em] = \Big(\dfrac{25 - 169}{144}\Big)\\[1em] = \Big(\dfrac{- 144}{144}\Big)\\[1em] = - 1$

Hence, option 4 is the correct option.