The identity we want to verify is:
[tex]\displaystyle{ \frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} =2\sec x[/tex].
Note that sec(x)=1/cosx, so we write the right hand side as 2/(cos(x)).
We multiply the first expression by (cosx)/(cosx), and the remaining two by (1+sinx)/(1+sinx) to have them in common numerators:
[tex]\displaystyle{ \frac{\cos^2x}{(\cos x)(1+\sin x)} + \frac{(1+\sin x)^2}{(\cos x)(1+\sin x)} = \frac{2(1+\sin x)}{(\cos x)(1+\sin x)} [/tex].
Now we can only consider the numerators. The right hand side, expanding (1+sinx)^2 becomes:
[tex]\cos^2x+1+2\sin x+\sin^2x
[/tex].
From the identity [tex]\sin ^2x+\cos^2x=1[/tex], the expression further simplifies to:
[tex]1+1+2\sin x=2+2\sin x=2(1+\sin x)[/tex]. This proves the identity.
