If $cos \theta = \frac{2}{3}$, then $2sec^2 \theta + 2 tan^2 \theta - 6 $ equals :

4 1 2 0

1

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