If $4-2 sin²θ-5 cosθ = 0$,$0° <θ< 90°$, then the value of $cosθ + tanθ$ is:

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

We are given that :-

4 - 2sin²θ - 5cosθ = 0

{ using , sin²θ  + cos²θ = 1 }

4 - 2 ( 1 - cos²θ ) - 5cosθ = 0

4 - 2 + 2cos²θ  - 5cosθ = 0

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,

cosθ + tanθ

= cos 60º + tan 60º

= \(\frac{1}{2}\) + √3

= \(\frac{1 +√3 }{2}\)