Find the value of θ, if $sec^2 θ + ( 1 - \sqrt{3}) tan θ - ( 1 + \sqrt{3}) = 0,$ where θ is an acute angle.
Answer & explanation
Correct answer: option 1
$sec^2 θ + ( 1 - \sqrt{3}) tan θ - ( 1 + \sqrt{3}) = 0,$
{ using , sec²θ - tan²θ = 1 }
= 1 + tan²θ + tanθ - √3 tanθ - 1 - √3
= tan²θ + tanθ - √3 tanθ - √3
= tanθ ( tanθ + 1 ) - √3 ( tanθ + 1 )
= ( tanθ + 1 ) ( tanθ -√3 )
tanθ = -1 ( not possible )
So, tanθ = √3
{ we know, tan60º = √3}
So, θ = 60º