If cot A + cosec A = 2 and A is an acute angle, then the value of $\frac{9 tan A+16 cosec A}{5sin A + 3 tan A}$ is :

4

cot A + cosec A = 2      ------(1)

we know, cosec²A - cot²A = 1 

( cosecA - cotA ) ( cosecA + cotA ) = 1

( cosecA - cotA ) × 2 = 1

( cosecA - cotA ) = \(\frac{1}{2}\)     ----(2)

On adding equation 1 and 2.

2 cosec A = 2 + \(\frac{1}{2}\)

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

By using  pythagoras theorem,

P² + B² = H²

4² + B² = 5²

B = 3

Now,

\(\frac{9tanA + 16 cosecA }{ 5sinA + 3 tanA }\)

= \(\frac{9×P/B  + 16×H/P}{ 5×P/H + 3×P/B }\)

= \(\frac{9× 4/3  + 16×5/4}{ 5×4/5 + 3×4/3}\)

= \(\frac{12  + 20}{ 4 + 4}\)

= \(\frac{32}{8}\)

= 4