Question

What is the value of $\frac{{(\sec \theta +1)}^2+{(1-\sec \theta )}^2}{1+{\sec }^2\theta }$ ?

Answer 1

$\frac{{(a+b)}^2+{(a-b)}^2}{a^2+b^2}=2$
In this case $a=\sec \theta$ b=1 Hence the value of the given expression is 2