What is the value of $\left(\frac{1+sec^2A}{1+cos^2A}\right)\left(\frac{1+sin^2A}{1+cosec^2A}\right)$?

$tan^2A$

$\left(\frac{1+sec^2A}{1+cos^2A}\right)\left(\frac{1+sin^2A}{1+cosec^2A}\right)$

Put A = 45°

= $\frac{1+2}{1+\frac{1}{2}}.\frac{1+\frac{1}{2}}{1+2}$

= 1

This is true for option 3 and 4

Solving, 

$\left(\frac{1+sec^2A}{1+cos^2A}\right)\left(\frac{1+sin^2A}{1+cosec^2A}\right)$

= $\frac{1+\frac{1}{cos^2A}}{1+cos^2A}.\frac{1+sin^2A}{1+\frac{1}{sin^2A}}$

= $\frac{sin^2A}{cos^2A}$ = tan²A