If A and B are acute angles and sec A = 3; cot B = 4, then the value of $\frac{cosec^2A +sin^2B}{cot^2+sec^2B}$ is :

$\frac{322}{323}$

With angle A,

secA = 3

P² + B² = H²

P² + 1² = 3²

P² = 8

P = 2√2

And with angle B

cotB = 4

P² + B² = H²

1² + 4² = H²

H² = 17

H = √17

Now,

$\frac{cosec^2A +sin^2B}{cot^2A+sec^2B}$

= $\frac{ (3/2√2)² + (1/√17)²}{(1/2√2)² +(√17/4)²}$

= \(\frac{161/136 }{19/16}\)

= \(\frac{322 }{323}\)