I am working on my homework and I am not sure if I am doing this correctly. Please let me know which one of these are correct, if any.

Find tan angle in QII, Sin angle=3/4 This is what I tried.

tan^2 angle +1= Sec^2 angle

tan^2 angle +1=(1/cos)^2 angle

(3/4)^2+ 1/cos^2angle=1

9/16+ 1/cos^2 angle=1

1/cos^2 angle=7/16

tan^2 angle=7/16

tan angle=+- squrt of 7/6, tan angle in Quad II= -squrt of 7/6.

also this one

tan^2 angle+1=sec^2 angle

tan^2 angle+1=(3/4)^2

tan^2 angle +1=(4/3)^2

tan^2 angle+1= 16/9

tan^2 angle= 7/9

tan angle = +- squrt of 7/9, tan angle = +- squrt of 7/3

tan angle in quad II= -squrt of 7/3

I would like to finish my homework but I do not want to do it all wrong. Please let me know if one of these ways is correct, if not could you please explain the right method to me so I can understand and do the rest correctly.

Thank you very much!

Keith Stevens

keithstevens@ctc.net