Proving Trigonometric Identities Prove: 1/ 1 + sinx = sec^2 x - tanxsecx

We need to prove
that:

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

We will start from the right side an prove the
right side.

==> sec^2 (x) -
tanx*secx.

We know that sec(x) =
1/cos(x)

and tanx =
sinx/cosx

==> sec^2 x - tanx*secx =
(1/cosx)^2 -sinx/cosx *
1/cosx

                                          = 1/cos^2
x  - sinx/cos^2 x

                                        =
(1-sinx) / cos^2 x

But we know that: sin^2 x + cos^2 x =
1

==> cos^2 x = 1- sin^2
x

==> (1-sinx)/cos^2 x = (1-sinx)/(1-sin^2
x)

Now we will factor the
denominator.

==> (1-sinx)/cos^2 x=
(1-sinx)/(1-sinx)(1+sinx)

Now we will reduce
similar.

==> sec^2 x - tanx*secx = 1/(1+sinx) .....
q.e.d