If $5 \cos ^2 \theta+1=3 \sin ^2 \theta, 0^{\circ}<\theta<90^{\circ}$, then what is the value of $\frac{\tan \theta+\sec \theta}{\cot \theta+{cosec} \theta}$.

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

5 cos²θ + 1 = 3 sin²θ

{ sin²θ + cos²θ = 1 }

5(1 - sin²θ) + 1 = 3 sin²θ

5 - 5sin²θ) + 1 = 3 sin²θ

8sin²θ = 6

sinθ  = \(\frac{√3}{2}\)

{ we know, sin60º  = \(\frac{√3}{2}\) }

So,  θ = 60º

Now,

\(\frac{ tanθ + secθ }{cotθ + cosecθ}\)

= \(\frac{ tan 60º + sec 60º }{cot 60º + cosec 60º}\) 

= \(\frac{ √3 + 2 }{1/√3 + 2/√3}\)

= \(\frac{ 3 + 2√3 }{3}\)