<0 and θ does not lie in second quadrant. Thus, θ lies is third quadrant. So, θ∕2 must be lie in second quadrant. There tanθ∕2 must be negative. then, cosθ=
3
5
=
B
H
∴tanθ=
4
3
So, tanθ=
2tanθ∕2
1−tan2θ∕2
let tanθ∕2=x ⇒
4
3
=
2x
1−x2
⇒4−4x2=6x ⇒2x2+3x−2=0 ⇒2x2+4x−x−2=0 ⇒(2x−1)(x+2)=0 Therefore, x=1∕2 or x=−2 ⇒tanθ∕2=1∕2 or tanθ∕2=−2 ∵tanθ∕2 is negative Hence, tanθ∕2=−2
