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

Feb 9, 2016

$- \frac{34}{16}$

#### Explanation:

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

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,$\sec x$= - $\sqrt{\frac{34}{9}}$ = - $\frac{\sqrt{34}}{3}$

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

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

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