How do you find sec 2x, given tan x = 5/3 and sin x< 0?

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How do you find sec 2x, given tan x = 5/3 and sin x< 0?

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Answer 1 · 2016-02-09T19:58:40

$- \frac{34}{16}$ $-34/16$

Explanation:

Starting from tan x, you can find sec x, because of the trigonometric identity 1 + ${tan}^{2} x$ $tan^2x$ = ${sec}^{2} x$ $sec^2x$
1+ ${(\frac{5}{3})}^{2}$ $(5/3)^2$ = ${sec}^{2} x$ $sec^2x$
$sec x$ $secx$ = $\sqrt{\frac{34}{9}}$ $sqrt(34/9)$

But since x is in Quadrant II, sec x has to be negative. That's because sec x has the same sign as cos x, because sec x = 1 / cos x. We know that cos x is negative is Quadrant II, therefore so is sec x. So, $sec x$ $secx$ = - $\sqrt{\frac{34}{9}}$ $sqrt(34/9)$ = - $\frac{\sqrt{34}}{3}$ $sqrt(34)/3$

Since sec x and cos x are reciprocals of each other,
cos x = 1/sec x = - $\frac{3}{\sqrt{34}}$ $3/sqrt(34)$
${cos}^{2} x =$ $cos^2x=$ 9/34.......... eq (i)

Now use the identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ to find sin x:
${sin}^{2} x$ $sin^2x$ = $1 - (\frac{9}{34})$ $1-(9/34)$ =25/34 ......eq (ii)
Again, we know that sin x is positive in Quadrant II

We know,
sec2x= $\frac{1}{cos (2 x)}$ $1/cos(2x)$
= $\frac{1}{{cos}^{2} x - {sin}^{2} x}$ $1/(cos^2x-Sin^2x)$
substituting the values,
sec2x= $\frac{1}{(\frac{9}{34}) - (\frac{25}{34})}$ $1/((9/34)-(25/34)$ = $- \frac{34}{16}$ $-34/16$

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How do you find sec 2x, given tan x = 5/3 and sin x< 0?

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