If cos θ=23, then 2 sec2 θ + 2 tan2 θ − 7 is equal to

Given that: $\cos \theta=\frac{2}{3}$

We have to find $2 \sec ^{2} \theta+2 \tan ^{2} \theta-7$

As we are given

$\cos \theta=\frac{2}{3}$

$\Rightarrow$ Base $=2$

$\Rightarrow$ Hypotenuse $=3$

$\Rightarrow$ Perpendicular $=\sqrt{(3)^{2}-(2)^{2}}$

$\Rightarrow$ Perpendicular $=\sqrt{5}$

We know that:

$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

$\tan \theta=\frac{\text { perpendicular }}{\text { Base }}$

Now we have to find: $2 \sec ^{2} \theta+2 \tan ^{2} \theta-7 .$ So

$2 \sec ^{2} \theta+2 \tan ^{2} \theta-7$

$=2\left(\frac{3}{2}\right)^{2}+2\left(\frac{\sqrt{5}}{2}\right)^{2}-7$

$=\frac{18}{4}+\frac{10}{4}-7$

$=\frac{18+10-28}{4}$

$=0$

Hence the correct option is $(b)$