Question

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?

Answer 1

`sec(theta) = (3sqrt(5))/4`

Explanation:

So, what we want to solve is this:

If `tan(theta) = sqrt(29)/4` ,what is `sec(theta)`?

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

`tan^2(theta) + 1 = sec^2(theta)`

If we solve this expression for `sec(theta):`

`sec(theta) = sqrt(tan^2(theta) + 1)`

Now, all we do is plug in the value we were given for `tan(theta)`:

`sec(theta) = sqrt((sqrt(29)/4)^2 + 1) = sqrt(29/16 + 1) = sqrt(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:

`=> sec(theta) = sqrt(45)/4`

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

`=> sec(theta) = (sqrt(9) * sqrt(5))/4 = (3sqrt(5))/4`

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

Hope that helped :)