If sec A$ =\frac{\sqrt{11}}{3}$, then the value of $\frac{cosec^2A + tan^2A}{sin^2+cot^2A}$ is :

$\frac{11}{9}$

secA =  \(\frac{√11 }{3}\)

{ cosx =  \(\frac{B }{H}\) }

P² + B² = H²

P² + 9 = 11

P = √2

Now,

 \(\frac{cosec²A + tan²A  }{sin²A + cot²A}\)

= \(\frac{ (√11/√2)² +(√2/3)²   }{(√2/√11)² + (3/√2)² }\)

=  \(\frac{11/2 + 2/9  }{2/11 + 9/2}\)

=  \(\frac{(99+4)/18  }{(4+99)22}\)

=   \(\frac{22  }{18}\)

=   \(\frac{11  }{9}\)