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If sec(theta)=5/3 and the terminal point determined by theta is in quadrant 4, then: (check all correct) sin(theta)=-2/5 cos(theta)=3/5 csc(theta)=-5/4 tan(theta)=4/3
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@ranga can you help
|dw:1391616145758:dw| Remember in 4th quadrant only cos theta and sec theta are positive all other T-ratios are negative.
Write -4K in place of 4k in the diagram. \[\csc \theta=\frac{ BA }{AC }=\frac{ 5K }{-4K }=-\frac{ 5 }{4 }\] \[\tan \theta=\frac{ AC }{BC }=\frac{ -4K }{3K }=-\frac{ 4 }{3 }\] |dw:1391616822900:dw|
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