M.C.Q (1 Marks)

Q: If $\cos\theta=\frac{2}{3},$ then $2\sec^2\theta+2\tan^2\theta-7$ 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 $\cos\theta=\frac{2}{3}$ $\Rightarrow\text{Base}=2$ $\Rightarrow\text{Hypotenuse}=3$ $\Rightarrow\text{Perpendicular}=\sqrt{(3)^2-(2)^2}$ $\Rightarrow\text{Perpendicular}=\sqrt{5}$ 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 $(b)$

Question

If $\cos	heta=\frac{2}{3},$ then $2\sec^2	heta+2	an^2	heta-7$ is equal to:

Vidyadip · Accepted Answer

Given that $\cos	heta=\frac{2}{3}$
We have to find $2\sec^2	heta+2	an^2	heta-7$
As we are given
$\cos	heta=\frac{2}{3}$
$\Rightarrow	ext{Base}=2$
$\Rightarrow	ext{Hypotenuse}=3$
$\Rightarrow	ext{Perpendicular}=\sqrt{(3)^2-(2)^2}$
$\Rightarrow	ext{Perpendicular}=\sqrt{5}$
We know that:
$\cos	heta=\frac{	ext{Base}}{	ext{Hypotenuse}}$
$	an	heta=\frac{	ext{Perpindicular}}{	ext{Base}}$
Now we have to find $2\sec^2	heta+2	an^2	heta-7$
So,
$2\sec^2	heta+2	an^2	heta-7$
$=2\Big(\frac{3}{2}\Big)^2+2\Big(\frac{\sqrt{5}}{2}\Big)^2-7$
$=\frac{18}{4}+\frac{10}{4}-7$
$=\frac{18+10-28}{4}$
$=0$
Hence the correct option is $(b)$

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