If 5 cos A = 3, the value of sec² A - tan² A is :

1. 1

2. -1

3. $\dfrac{3}{4}$

4. $\dfrac{4}{3}$

Given:

5 cos A = 3

⇒ cos A = $\dfrac{3}{5}$

i.e., $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{3}{5}$

∴ If length of AB = 3x unit, length of AC = 5x unit.

In Δ ABC,

AC² = BC² + AB²

⇒ (5x)² = BC² + (3x)²

⇒ 25x² = BC² + 9x²

⇒ BC² = 25x² - 9x²

⇒ BC² = 16x²

⇒ BC = $\sqrt{16\text{x}^2}$

⇒ BC = 4x

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

= $\dfrac{AC}{BA} = \dfrac{5x}{3x} = \dfrac{5}{3}$

tan A = $\dfrac{Perpendicular}{Base}$

= $\dfrac{BC}{BA} = \dfrac{4x}{3x} = \dfrac{4}{3}$

Now, sec² A - tan² A

$= \Big(\dfrac{5}{3}\Big)^2 - \Big(\dfrac{4}{3}\Big)^2\\[1em] = \dfrac{25}{9} - \dfrac{16}{9}\\[1em] = \dfrac{25 - 16}{9}\\[1em] = \dfrac{9}{9}\\[1em] = 1$

Hence, option 1 is the correct option.