MCQ
If $\cos\theta=\frac{2}{3},$ then $2\sec^2\theta+2\tan^2\theta-7$ is equal to:
- A$1$
- ✓$0$
- C$3$
- D$4$
✓
Answer
Correct option: B.
$0$
Given that $\cos\theta=\frac{2}{3}$
We have to find $2\sec^2\theta+2\tan^2\theta-7$
As we are given
$\cos\theta=\frac{2}{3}$
$\Rightarrow\text{Base}=2$
$\Rightarrow\text{Hypotenuse}=3$
$\Rightarrow\text{Perpendicular}=\sqrt{(3)^2-(2)^2}$
$\Rightarrow\text{Perpendicular}=\sqrt{5}$
We know that:
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$
$\tan\theta=\frac{\text{Perpindicular}}{\text{Base}}$
Now we have to find $2\sec^2\theta+2\tan^2\theta-7$
So,
$2\sec^2\theta+2\tan^2\theta-7$
$=2\Big(\frac{3}{2}\Big)^2+2\Big(\frac{\sqrt{5}}{2}\Big)^2-7$
$=\frac{18}{4}+\frac{10}{4}-7$
$=\frac{18+10-28}{4}$
$=0$
Hence the correct option is $(b)$
We have to find $2\sec^2\theta+2\tan^2\theta-7$
As we are given
$\cos\theta=\frac{2}{3}$
$\Rightarrow\text{Base}=2$
$\Rightarrow\text{Hypotenuse}=3$
$\Rightarrow\text{Perpendicular}=\sqrt{(3)^2-(2)^2}$
$\Rightarrow\text{Perpendicular}=\sqrt{5}$
We know that:
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$
$\tan\theta=\frac{\text{Perpindicular}}{\text{Base}}$
Now we have to find $2\sec^2\theta+2\tan^2\theta-7$
So,
$2\sec^2\theta+2\tan^2\theta-7$
$=2\Big(\frac{3}{2}\Big)^2+2\Big(\frac{\sqrt{5}}{2}\Big)^2-7$
$=\frac{18}{4}+\frac{10}{4}-7$
$=\frac{18+10-28}{4}$
$=0$
Hence the correct option is $(b)$
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