SOLUTION: Can you help me find csc0 if cot0 = the - square root of 3 and cos 0 <0 i tried to do cot^20 = (- square root of 3)^2 cot2x= -3 1/tan = -3 1(1/1-sec2x) = (-3)1 1-sec^2x = -3

Question 835223: Can you help me find csc0 if cot0 = the - square root of 3 and cos 0 <0
i tried to do cot^20 = (- square root of 3)^2
cot2x= -3
1/tan = -3
1(1/1-sec2x) = (-3)1
1-sec^2x = -3
1-1/cosx = -3
-1/cosx = -2
1(1/cosx) = (-2)1
cosx= -2
But I ended up confusing myself more than getting the answer right. Please help?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
find csc0 if cot0 = -sqrt(3) and cos 0 < 0
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Note: cot and cos are both negative in QII where x < 0 and y > 0
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Since cot = x/y, x = -sqrt(3) and y = 1
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Then r = sqrt[(sqrt(3))^2+1^2] = sqrt[3+1] = 2
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Your Problem:
csc(t) = r/y = 2/1 = 2
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Cheers,
Stan H.
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