Question

cos theta=2/3 find the value 2sec square theta + 2tan square theta -9
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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The value of the expression

$2\sec^2\theta+2\tan^2\theta-9=-2$

Step-by-step explanation:

Given :

The value of

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

To Find :

The value of

$2\sec^2\theta+2\tan^2\theta-9$

Solution :

Since we are given that The value of

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

and we have to calculate the value of

$2\sec^2\theta+2\tan^2\theta-9$

Since we know that

$\sec \theta=\frac{1}{\cos\theta}=\frac{1}{\frac{2}{3}}=\frac{3}{2}$

therefore $\sec^2\theta=(\frac{3}{2})^2=\frac{9}{4}$

and

$\tan^2\theta=\sec^2\theta-1$

$\tan^2\theta=(\frac{3}{2})^2-1=\frac{9}{4}-1=\frac{9-4}{4}$

$\tan^2\theta=\frac{5}{4}$

Now substituting these values in the expression

$2\sec^2\theta+2\tan^2\theta-9$

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

$=\frac{9}{2}+\frac{5}{2}-9$

$=\frac{14}{2}-9$

$=7-9$

$2\sec^2\theta+2\tan^2\theta-9=-2$

Hence we have calculated the value of the expression

   $2\sec^2\theta+2\tan^2\theta-9=-2$

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