If $\cos \theta=\frac{2}{3}$, then $2 \sec ^{2} \theta+2 \tan ^{2} \theta-7$ is equal to
(a) 1
(b) 0
(c) 3
(d) 4
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 $(b)$
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