If $2 \cos ^2 \theta-5 \cos \theta+2=0,0^{\circ}<\theta<90^{\circ}$, then the value of $(\sec \theta+\tan \theta)$ is:

$2+\sqrt{3}$

We are given :-

2cos²θ - 5cosθ + 2 = 0

2cos²θ  - 4cosθ  - cosθ  + 2 = 0

 2cosθ ( cosθ  - 2 ) - 1 ( cosθ  - 2 ) = 0

( 2cosθ  - 1 ) . ( cosθ  - 2 ) = 0

Either ( 2cosθ  - 1 )  = 0 or ( cosθ  - 2 ) = 0

( cosθ  - 2 ) = 0 is not possible

So, ( 2cosθ  - 1 )  = 0 

cosθ = \(\frac{1}{2}\)

{ We know, cos60º = \(\frac{1}{2}\) }

So, θ = 60º

Now,

secθ + tanθ

= sec60º + tan 60º

= 2 + √3