# SOLUTION: Evaluate: Sec (-7pi/6) + Csc (pi/3) + tan 8pi/3 This is what I have so far: Sec -210 = -2 times the square root of 3 / 3 Csc 30 = 2 tan 480 = - the square root of 3

Algebra ->  Algebra  -> Trigonometry-basics -> SOLUTION: Evaluate: Sec (-7pi/6) + Csc (pi/3) + tan 8pi/3 This is what I have so far: Sec -210 = -2 times the square root of 3 / 3 Csc 30 = 2 tan 480 = - the square root of 3       Log On

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 Question 509195: Evaluate: Sec (-7pi/6) + Csc (pi/3) + tan 8pi/3 This is what I have so far: Sec -210 = -2 times the square root of 3 / 3 Csc 30 = 2 tan 480 = - the square root of 3 What I got was -5 times the square root of 3 + 6. How do you get a - square root of 3? Answer by solver91311(16885)   (Show Source): You can put this solution on YOUR website! Let's look at the unit circle: Note that And recall that So, find on the unit circle, note that the -coordinate is , take the reciprocal and rationalize to get: Which is what you had. Now, recall that So, find on the unit circle, note that the -coordinate is , take the reciprocal and rationalize to get: Hence the sum of your first two terms is zero. Now for the third term: Note that Since the tangent function has a periodicity of , we can conclude that: Now, recall that So, find on the unit circle, note that the -coordinate is , the -coordinate is , and then take the quotient of the -coordinate divided by the -coordinate to get: John My calculator said it, I believe it, that settles it