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How do you use the half angle formula to evaluate #Tan (-195) #?

Answer 1

#sqrt3 - 2#

tan (-195) = - tan (195). Evaluate tan (195). Apply the trig identity: #tan 2a = (2tan a)/(1 - tan^2 a)# tan ((2(195)) = tan 390 = tan (30 + 360) = tan 30 = 1/sqrt3 Call tan (195) = t, we get: #1/sqrt3 = (2t)/(1 - t^2)#. After cross multiplication, we get: 1 - t^2 = 2sqrt3t t^2 + 2sqrt3t - 1 = 0. Solve this quadratic equation for t. D = d^2 = b^2 - 4ac = 12 + 4 = 16 --> d = +- 4 There are 2 real roots: #t = tan 195 = -(2sqrt3)/2 +- 4/2 = -sqrt3 +- 2# Therefor, #tan (-195) = - tan (195) = sqr3 +- 2# Since the arc (-195) is in Quadrant II, its tan is negative, then the negative answer is accepted. #tan (-195) = sqrt3 - 2.#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.