If $12 \cos^2 \theta - 2 \sin^2 \theta + 3\cos \theta = 3, 0^\circ < \theta < 90^\circ$, then what is the value of $\frac{cosec~\theta + \sec \theta}{\tan \theta + \cot \theta}$?

$\frac{1 + \sqrt{3}}{2}$

12 cos²θ - 2sin²θ + 3cosθ  = 3

12 cos²θ - 2 ( 1 - cos²θ ) + 3cosθ  = 3

14 cos²θ  + 3cosθ  - 5 = 0

14 cos²θ  + 10cosθ -7cosθ - 5 = 0

2cosθ ( 7cosθ + 5 ) - 1 ( 7cosθ + 5 ) = 0

( 2cosθ - 1 ) . ( 7cosθ + 5 ) = 0

Either ( 2cosθ - 1 ) = 0 Or  ( 7cosθ + 5 ) = 0 

( 7cosθ + 5 ) = 0 

cosθ = - \(\frac{5}{7}\)  ( not possible )

So, ( 2cosθ - 1 ) = 0

cosθ = \(\frac{1}{2}\)

{ we know, cos60º = \(\frac{1}{2}\) }

So , θ = 60º

Now,

\(\frac{cosecθ+ secθ }{tanθ + cotθ}\)

= \(\frac{cosec60º+ sec60º }{tan60º + cot60º}\)

= \(\frac{2/√3 +2  }{√3 + 1/√3}\)

= \(\frac{2√3 +2  }{4}\)

= \(\frac{√3 +1  }{2}\)