Precalculus(NEED HELP ASAP PLEASE!!)
posted by jh .
cot(theta)= 3
pi < theta < 3pi/2
Find:
sin(theta)= 1 ?
cos(theta)= 3 ?
tan(theta)= 1/3 ?
sec(theta)= 1/3 ?
csc(theta)= 1 ?
That's what I came up with, but they are not correct, PLEASE help me where I went wrong!!

Precalculus(NEED HELP ASAP PLEASE!!) 
bobpursley
tan theta= 1/3 which means sintheta 1/sqr10
draw the triangle Notice it is in the third quadrant, where cosine is negative. 
Precalculus(NEED HELP ASAP PLEASE!!) 
Damon
You do not seem to have calculated the hypotenuse which you need for sin and cos and sec and csc
sqrt (1^2+3^2) = sqrt 10 
Precalculus(NEED HELP ASAP PLEASE!!) 
jh
So,
tan(theta)=sin(theta)/cos(theta)
cos(theta)=sin(theta)/tan(theta)
=(1/sqrt(10))/(1/3)
= 3/sqrt(10) ??
But its incorrect!! What am I doing wrong? 
Precalculus(NEED HELP ASAP PLEASE!!) 
Josie
If sin t=1/4, find sin t + 6pi

Precalculus(NEED HELP ASAP PLEASE!!) 
Kayla
tan theta= 5 divided by 12 and theta is in quadrant 3 what does sin 2 theta equal
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