Trigonometric Ratios and Identities

∵cos‌θ=−‌

3

5

<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