(iv) tan² θ = 3 (sec θ – 1)

(sec² θ – 1) = 3 sec θ – 3

sec² θ – 1 – 3 sec θ + 3 = 0

sec² θ – 3 sec θ + 2 = 0

sec² θ – 2 sec θ – sec θ + 2 = 0

sec θ (sec θ – 2) – 1 (sec θ = 2) = 0

(sec θ – 1) (sec θ – 2) = 0

So, either sec θ – 1 = 0 or sec θ – 2 = 0

If sec θ – 1 = 0

sec θ = 1

⇒ θ = 0^o

And, if sec θ – 2 = 0

sec θ = 2

⇒ θ = 60^o

Thus, θ = 0^o or 60^o