Answer
$(tan~x+1)^2~cos~x=2~sin~x+sec~x$
Work Step by Step
$1+tan^2x=sec^2x=\frac{1}{cos^2x}$
$tan~x=\frac{sin~x}{cos~x}$
$(tan~x+1)^2~cos~x=(tan^2x+2~tan~x+1)~cos~x=(2~tan~x+sec^2x)~cos~x=(2~\frac{sin~x}{cos~x}+\frac{1}{cos^2x})~cos~x=2~sin~x+\frac{1}{cos~x}=2~sin~x+sec~x$
