Question

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

Answer 1

`sqrt3 - 2`

Explanation:

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.`