## dontothabonbon 2 years ago Let θ be an angle in quadrant II such that sin

1. cwrw238

|dw:1372535299105:dw|

2. cwrw238

can you continue from here?

3. dontothabonbon

Thanks. I'm not sure what to do from here.

4. cwrw238

tan x = opposite / adjacent = -1 / sqrt15 sec x = hypotenuse / adjacent = ?

5. jdoe0001
6. jdoe0001

|dw:1372535888225:dw|

7. dontothabonbon

Alright now I see. Thanks to both of you.

8. cwrw238

yw

9. dontothabonbon

Would the hypotenuse become a negative when calculating sec? Or would it just be: sec x = 4/sqrt15?

10. dontothabonbon

Or better yet, why did the 1 become negative when calculating the tangent?

11. Jhannybean

Because your function is in the second quadrant where sine is positive and cosine is negative.

12. cwrw238

the whole ratio became negative tangent : opposite = 1 , adjacent = - sqrt15 tan = 1 / -sqrt15 = -1/sqrt15

13. Jhannybean

$\large \left(-\frac{1}{\sqrt{15}}\right)$

14. dontothabonbon

So, sec = -4/sqrt15?

15. cwrw238

yes

16. dontothabonbon

ahh okay I see now thanks.

17. cwrw238

x values to the left of y-axis are negative y values below x axis are also negative