# Find secant of theta if tangent of theta is the square root of 29 over 4?

## If you use the pythagorean identities, of tangent ^2 +1 = secant ^2, how do you solve?

Apr 7, 2018

$\sec \left(\theta\right) = \frac{3 \sqrt{5}}{4}$

#### Explanation:

So, what we want to solve is this:

If $\tan \left(\theta\right) = \frac{\sqrt{29}}{4}$ ,what is $\sec \left(\theta\right)$?

There's a number of ways to solve this problem, but let's solve it using a trigonometric identity:

${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$

If we solve this expression for $\sec \left(\theta\right) :$

$\sec \left(\theta\right) = \sqrt{{\tan}^{2} \left(\theta\right) + 1}$

Now, all we do is plug in the value we were given for $\tan \left(\theta\right)$:

$\sec \left(\theta\right) = \sqrt{{\left(\frac{\sqrt{29}}{4}\right)}^{2} + 1} = \sqrt{\frac{29}{16} + 1} = \sqrt{\frac{45}{16}}$

..and that's your answer! However, we can polish this up a little so it looks nicer. For one, we can remove the radical in the denominator, since 16 is a perfect square:

$\implies \sec \left(\theta\right) = \frac{\sqrt{45}}{4}$

Now, we also know that $45 = 9 \cdot 5$. Notice that 9 is a perfect square, so we can further simplify:

$\implies \sec \left(\theta\right) = \frac{\sqrt{9} \cdot \sqrt{5}}{4} = \frac{3 \sqrt{5}}{4}$

And that is pretty much as far as we can simplify. So, this would be your final answer.

Hope that helped :)