Question 581265
W(t) is in Quadrant II and sec(t) = –sqrt(10). Find tan(t).
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sec=hypotenuse/adjacent side=-√10/1
Working with a right triangle in quadrant II where tan<0
hypotenuse=&#8730;10
adjacent side=-1
By Pythagorean Theorem
opposite side=&#8730;10-1^2)=&#8730;9=3
tan(t)=opposite/adjacent=3/-1=-3
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You could also use the following trig identity to get the same answer:
tan^2+1=sec^2
tan^2=sec^2-1=&#8730;10)^2-1=10-1=9
tan=-3 (in quadrant II)