sin2x+cos2x=1 Given sinx=3/5,cosx>0. Then
cosx=1−sin2x=1−(3/5)2=4/5 Given cosy=−4/5,siny>0. Then
siny=1−cos2y=1−(−4/5)2=3/5
a)
cotx=sinxcosx=3/54/5=4/3
cosecy=siny1=3/51=5/3
cotx+cosecy=4/3+5/3=3
b)
sin2x=2sinxcosx=2(3/5)(4/5)=24/25
tany=cosysiny=−4/53/5=−3/4
tan2y=1−tan2y2tany=1−(−3/4)22(−3/4)=−24/7
sin2x+tan2y=24/25−24/7=−432/175
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