If $2 \cos ^2 \theta=3 \sin \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\left(\sec ^2 \theta-\tan ^2 \theta+\cos ^2 \theta\right)$ is:

$\frac{7}{4}$

2 cos²θ = 3 sinθ

Let us assume that ,

θ= 30º

2 cos²30º = 3 sin30º

2 × \(\frac{3}{4}\)  = 3 × \(\frac{1}{2}\)

\(\frac{3}{2}\)  = \(\frac{3}{2}\)

LHS = RHS  ( satisfied )

So, θ = 30º

Now,

( sec²θ - tan²θ + cos²θ )

= ( sec²30º - tan²30º + cos²30º ) 

= \(\frac{4}{3}\) - \(\frac{1}{3}\) + \(\frac{3}{4}\)

= \(\frac{3}{3}\) + \(\frac{3}{4}\)

= \(\frac{7}{4}\)