If $4{\sin }^2\theta =3(1+\cos \theta ),0°<\theta <90°,$ then what is the value of $\sqrt{15}\tan \theta +\frac{4}{\sqrt{15}}\sin \theta +2\sec \theta$

The correct answer is 24.

We are given the trigonometric equation:
$4{\sin }^2\theta =3(1+\cos \theta )$
where $0°<\theta <90°$.

First, we can use the fundamental trigonometric identity ${\sin }^2\theta =1-{\cos }^2\theta$ to rewrite the equation entirely in terms of $\cos \theta$:
$4(1-{\cos }^2\theta )=3(1+\cos \theta )$

We know that $1-{\cos }^2\theta$ can be factored as a difference of squares: $(1-\cos \theta )(1+\cos \theta )$. Substituting this back into the equation:
$4(1-\cos \theta )(1+\cos \theta )=3(1+\cos \theta )$

Since $0°<\theta <90°$, we have $0<\cos \theta <1$, which means $1+\cos \theta \ne 0$. Thus, we can divide both sides of the equation by $1+\cos \theta$:
$4(1-\cos \theta )=3$

Now, we solve for $\cos \theta$:
$1-\cos \theta =\frac{3}{4}$
$\cos \theta =1-\frac{3}{4}=\frac{1}{4}$

Since $\cos \theta =\frac{1}{4}$, we can determine the other trigonometric ratios for the acute angle $\theta$:
$\sec \theta =\frac{1}{\cos \theta }=4$
$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-{\left(\frac{1}{4}\right)}^2}=\sqrt{1-\frac{1}{16}}=\sqrt{\frac{15}{16}}=\frac{\sqrt{15}}{4}$
$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\sqrt{15}/4}{1/4}=\sqrt{15}$

Now, we substitute these values into the expression we wish to evaluate:
$E=\sqrt{15}\tan \theta +\frac{4}{\sqrt{15}}\sin \theta +2\sec \theta$
$E=\sqrt{15}\left(\sqrt{15}\right)+\frac{4}{\sqrt{15}}\left(\frac{\sqrt{15}}{4}\right)+2\left(4\right)$

Simplifying each term step-by-step:
1. The first term: $\sqrt{15}\times \sqrt{15}=15$
2. The second term: $\frac{4}{\sqrt{15}}\times \frac{\sqrt{15}}{4}=1$
3. The third term: $2\times 4=8$

Adding them together:
$E=15+1+8=24$

Therefore, the value of the given expression is 24.