Given tantheta=-3/4tanθ=−34 and 90<theta<18090<θ<180, how do you find tan(theta/2)tan(θ2)?

1 Answer
Nghi N.
Sep 26, 2016

tan (t/2) = 3tan(t2)=3

Explanation:

Call tan (t/2) = x and use trig identity:
tan 2t = (2tan t)/(1 - tan^2 t)tan2t=2tant1−tan2t
tan 2x = (-3/4) = (2x)/(1 - x^2)tan2x=(−34)=2x1−x2
Cross multiply:
3x^2 - 3 = 8x3x2−3=8x
Solve the quadratic equation for x.
3x^2 - 8x - 3 = 03x2−8x−3=0
D = d^2 = b^2 - 4ac = 64 + 36 = 100D=d2=b2−4ac=64+36=100 --> d = +- 10d=±10
There are 2 real roots:
x = - b/(2a) +- d/(2a) = 8/6 +- 10/6 = (4 +- 5)/3x=−b2a±d2a=86±106=4±53
x1 = 3, and x2 = - 1/3x2=−13
tan (t/2) = x1 = 3tan(t2)=x1=3
and tan (t/2) = x2 = - 1/3tan(t2)=x2=−13
Because t is in Quadrant II, then (t/2)(t2) is in Quadrant I, and tan (t/2)tan(t2) is positive.
There for: tan (t/2) = 3tan(t2)=3

Related questions
See all questions in Half-Angle Identities
Impact of this question
5428 views around the world
You can reuse this answer
Creative Commons License