If $3 \sec^{2} \theta + \tan \theta - 7 = 0$, $0^{\circ} < \theta < 90^{\circ}$,then what is the value of $\left(\frac{2 \sin \theta + 3 cos \theta}{cosec~ \theta + \sec \theta}\right)$ ?

$\frac{5}{4}$

3sec²θ + tanθ - 7 = 0

Put θ = 45º

3 × 2 + 1 - 7

= 0

LHS = RHS ( satisfied with angle 45º )

Now , \(\frac{2sinθ + 3cosθ }{cosecθ + secθ}\) 

= \(\frac{2× 1/√2  + 3× 1/√2 }{√2 + √2}\)

= \(\frac{2 + 3 }{2×√2×√2 }\) 

= \(\frac{5}{4}\)