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 :
Answer & explanation
Correct answer: option 2
$\frac{cosec^2θ}{cosec^245-cot^245} =\frac{13}{4}$
{ 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,
$\frac{52cos^2θ-9tan^2θ}{18sec^2θ+8cot^2θ}$
= \(\frac{52×3/13 - 9×4/9}{18×13/9 + 8 × 9/4}\)
= \(\frac{32}{44}\)
= \(\frac{8}{11}\)