The correct option is A 2
Given:
(1−tanθ)(1+tanθ)sec2θ+2tan2θ=0
⇒(1−tan2θ)(1+tan2θ)+2tan2θ=0
Let tan2θ=t
⇒(1−t)(1+t)+2t=0⇒1−t2+2t=0⇒2t=t2−1
As tan2θ∈W
So, by observation, t=3 is the only solution.
∵23=32−1
Now,
tan2θ=3
⇒tanθ=±√3
As θ∈(−π2,π2), so
θ=±π3
Hence, the number of solution is 2.