How do you find the values of the six trigonometric functions given #sintheta=3/5# and #theta# lies in Quadrant II?

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Answer 1 · 2017-02-28T21:08:51

See below

#sintheta=3/5# and #theta# is in Quadrant #"II"#

This means #sin(180-theta)=3/5#

So in a right-angled triangle, the #"opp"=3# and the #"hyp"=5#. Thus the #"adj"=sqrt(5^2-3^2)=sqrt16=4#

Therefore, #cos(180-theta)=4/5#

#cos180costheta+sin180sintheta=4/5#

#-costheta=4/5#

#costheta=-4/5#

#tantheta=sintheta/costheta=(3/5)/(-4/5)=-3/4#

#csctheta=1/sintheta=1/(3/5)=5/3#

#sectheta=1/costheta=1/(-4/5)=-5/4#

#cottheta=1/tantheta=1/(-3/4)=-4/3#