If cos θ = 2/3, then 2sec^2 θ + 2 tan^2 θ - 6 equals :

cosθ = \(\frac{2}{3}\)

{ we know, cosθ = \(\frac{B}{P}\) }

By using pythagoras thorem,

P² + B² = H²

P² + 2² = 3²

P = \(\sqrt { 5}\)

Now,

2 sec²θ + 2tan²θ - 6

= 2(\(\frac{9}{4}\))  + 2(\(\frac{5}{4}\)) - 6 

= (\(\frac{9}{2}\))  + (\(\frac{5}{2}\)) - 6

= 1