If 0∘ < θ < 90∘, and tan2θ-(√3+1) tan θ+√3=0, then θ=?

(1)30^∘  (2) 45^∘ (3) 60^∘ (4) 45^∘ or 60^∘

Solution:

tan²θ- √3 tan θ -tan θ +√3=0

= tan θ (tan θ -√3) -1(tan θ -√3) =0

= (tan θ -1) (tan θ -√3)=0

= tan θ =1=tan 45^∘

= θ = 45^∘

Again,

tan θ -√3 =0

⇒ tan θ =√3 = tan 60^∘

⇒ θ = 60^∘

Then the option is (3)