# Math Help - Secant, and cosecant angles

1. ## Secant, and cosecant angles

Hi!
I have trouble dealing with secΘ and cotΘ.

1. If sinΘ = -1÷ √2 and 270° < Θ < 360°, find the values of secΘ and cotΘ
2. If cosΘ = -24÷ 25 and sinΘ<0, find the values of secΘ and cotΘ

I know that -1÷ √2 can be -45° or 225° or 585°, but it doesn't fall under the condition 270° < Θ < 360°. Then I got stuck.....

2. ## Re: Secant, and cosecant angles

$\sin\theta=\frac{-1}{\sqrt{2}}$

Taking $\theta=-45$ and knowing that $\cos(-x)=\cos(x)$ makes this fairly straightforward once you've used the unit circle.

For $\cos\theta =\frac{-24}{25}$, consider a $7$, $24$, $25$ right triangle and the property $-\sin\theta=\sin{(-\theta)}$ combined with the one mentioned previously. I think this is all you should need to get a solution.

3. ## Re: Secant, and cosecant angles

It would help if I'd read the OP carefully before replying. Corrected my previous response.

4. ## Re: Secant, and cosecant angles

Originally Posted by Quacky
$\sin\theta=\frac{-1}{\sqrt{2}}$

Taking $\theta=-45$ and knowing that $\cos(-x)=\cos(x)$ makes this fairly straightforward once you've used the unit circle.

For $\cos\theta =\frac{-24}{25}$, consider a $7$, $24$, $25$ right triangle and the property $-\sin\theta=\sin{(-\theta)}$ combined with the one mentioned previously. I think this is all you should need to get a solution.
Thank you so much for your help!!

And this is what I got.....
1. Sec(-1÷ √2) = √2
Cot(-1÷ √2) = -1

2. Sec(-24÷ 25) ~ -1.04
Cot(-24÷ 25) ~ 3.4