Prove (tanx+1)sqcosx= secx +2sinx

(tanx+1)sqcosx= secx +2sinx please prove that the left side equal the right side. I've been working for hours but couldn't find the right answer.
Thank you for your help and 10 points for the correct answer.
Note: (tanx+1) is squared then multiplied by cosx

think of it like this

(tanx+1)squared is the same as (tanx+1)(tanx+1)

so (tanx+1)(tanx+1)cosx = secx+2sinx

then foil (tanx+1) and (tanx +1)

{tan^2(x) + tanx + tanx + 1} cosx = secx + 2sinx

tan^2(x)+1 = sec^2(x) **this is one of the Pythagorean identities**

now distribute the cosx (also note tanx + tanx becomes 2tanx)

cosx(sec^2(x)) + 2cosx(tanx) = secx +2sinx

note secx = 1/cosx and tanx = sinx/cosx

cosx/cos^2(x) + 2cosxsinx/cosx = secx + 2sinx

1/cosx + 2sinx = secx + 2sinx

secx + 2sinx = secx + 2sinx

***tan^2(x) means (tanx) squared ***

(1+tanx)^2*cosx
=(1+tan^2(x)+2tanx)*cosx
= (sec^2(x)+2tanx)*cosx
= (secx/cosx + 2sinx/cosx)*cosx
= secx + 2sinx