If $\frac{cosec^2θ}{cosec^245-cot^245} =\frac{13}{4}, 0° < θ < 90°$, then the value of $\frac{52cos^2θ-9tan^2θ}{18sec^2θ+8cot^2θ}$ will be :

$\frac{8}{11}$

$\frac{cosec^2θ}{cosec^245-cot^245} =\frac{13}{4}$

{ cosec² A - cot² A = 1 }

$\frac{cosec^2θ}{1} =\frac{13}{4}$

{ cosec θ = \(\frac{H}{P}\) }

By using pythagoras theorem,

P² + B² = H²

4 + B² = 13

B = 3

Now,

$\frac{52cos^2θ-9tan^2θ}{18sec^2θ+8cot^2θ}$

= \(\frac{52×3/13 - 9×4/9}{18×13/9 + 8 × 9/4}\)

= \(\frac{32}{44}\) 

= \(\frac{8}{11}\)