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

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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

We take the RHS: sec^2x-tanxsecx = 1/cos^2x - sinxcos^2x. as tanx = sinx/cosx.

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

sec^2x-tanxsecx = (1-sin^2)/(cos^2x)(1+sinx)

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

sec^2x-tanxsecx = 1/(1+sinx) = LHS.

Therefore

1/ 1 + sinx = sec^2 x - tanxsecx

Therefore 1/(1 + sinx) = sec^2 x - tanxsecx.

L:H:S ≡ 1 / (1 + sinx)

= 1 (1 - sinx) / (1 + sinx) (1 - sinx)

= (1 - sinx) / (1 - sin²x)

*⇒ use 1 - sin²A = cos²A*

= (1 - sinx) / cos²x

= 1/cos²x - sinx/cos²x

= sec²x - tanx/cosx

= sec²x - tanx.secx

L:H:S ≡ R:H:S