Given that sin(theta) = 3/5 and theta is in quadrant II, we can use the Pythagorean identity sin^2(theta) + cos^2(theta) = 1 to find the value of cos(theta).
sin^2(theta) + cos^2(theta) = 1
(3/5)^2 + cos^2(theta) = 1
9/25 + cos^2(theta) = 1
cos^2(theta) = 1 - 9/25
cos^2(theta) = 16/25
cos(theta) = ±4/5 (since theta is in quadrant II)
Since theta is in quadrant II, cos(theta) is negative. Therefore, cos(theta) = -4/5.
Now, we can find the values of the remaining trigonometric functions:
tan(theta) = sin(theta) / cos(theta) = (3/5) / (-4/5) = -3/4
csc(theta) = 1/sin(theta) = 1 / (3/5) = 5/3
sec(theta) = 1/cos(theta) = 1 / (-4/5) = -5/4
cot(theta) = 1/tan(theta) = 1 / (-3/4) = -4/3
Therefore, the exact values of the remaining trigonometric functions of theta are:
tan(theta) = -3/4
csc(theta) = 5/3
sec(theta) = -5/4
cot(theta) = -4/3
find the exact value of each of the remaining trigonometric functions of theta. sintheta=3/5, theta in quadrant II
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