If $4 - 2 \sin^2 \theta - 5 \cos \theta = 0, 0^\circ < \theta < 90^\circ$, then the value of $\cos \theta - \tan \theta$ is:

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

We are given that ,

4 - 2 sin²θ - 5 cosθ= 0

{ using sin²θ + cos²θ = 1 }

4 - 2 ( 1 - cos²θ ) - 5 cosθ= 0

2cos²θ - 5 cosθ + 2 = 0

2cos²θ - 4 cosθ - cosθ + 2 = 0

2 cosθ ( cosθ - 2 ) - 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

= cos60º - tan60º

= \(\frac{1}{2}\) - \(\sqrt { 3}\)

= \(\frac{1 - 2√3}{2}\)