If 3cos θ = 2 then (2sec2 θ + 2tan2 θ – 7) = ?

Given : $3 \cos \theta=2$

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

Since, $\cos \theta=\frac{B}{H}$

$\Rightarrow B=2$ and $H=3$

Using Pythagoras theorem,

$P^{2}+B^{2}=H^{2}$

$\Rightarrow P^{2}+2^{2}=3^{2}$

$\Rightarrow P^{2}=9-4$

$\Rightarrow P^{2}=5$

$\Rightarrow P=\sqrt{5}$

Therefore,

$\sec \theta=\frac{H}{B}=\frac{3}{2}$

$\tan \theta=\frac{P}{B}=\frac{\sqrt{5}}{2}$

Now,

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

$=\left(2\left(\frac{9}{4}\right)+2\left(\frac{5}{4}\right)-7\right)$

$=\left(\frac{9}{2}+\frac{5}{2}-7\right)$

$=\left(\frac{9+5}{2}-7\right)$

$=\left(\frac{14}{2}-7\right)$

$=(7-7)$

$=0$

Hence, the correct option is (a).