If cos θ = $\Big(\dfrac{2}{3}\Big)$, then 2 sec²θ + 2 tan²θ − 7 is equal to:

1. 0

2. 1

3. 3

4. 4

cos θ = $\dfrac{2}{3}$

sec θ = $\dfrac{3}{2}$

sec² θ = $\dfrac{9}{4}$

tan² θ = sec² θ - 1 = $\dfrac{9}{4} - 1 = \dfrac{5}{4}$

Given,

⇒ 2 sec²θ + 2 tan²θ − 7

$\Rightarrow 2\Big(\dfrac{9}{4}\Big) + 2\Big(\dfrac{5}{4}\Big) - 7 \\[1em] \Rightarrow \Big(\dfrac{18}{4}\Big) + \Big(\dfrac{10}{4}\Big) - 7 \\[1em] \Rightarrow \Big(\dfrac{28}{4}\Big) - 7 \\[1em] \Rightarrow 7 - 7 \\[1em] \Rightarrow 0.$

Hence, option 1 is the correct option.