We have: frac(1 + sin(x))(1 - sin(x)) = (sec(x) + tan(x))^(2)1+sin(x)1−sin(x)=(sec(x)+tan(x))2
We will begin the proof from the right-hand side of the identity.
First, let's expand the parentheses:
= sec^(2)(x) + 2 sec(x) tan(x) + tan^(2)(x)=sec2(x)+2sec(x)tan(x)+tan2(x)
Then, let's apply the standard trigonometric identities sec(x) = frac(1)(cos(x))sec(x)=1cos(x) and tan(x) = frac(sin(x))(cos(x))tan(x)=sin(x)cos(x):
= frac(1)(cos^(2)(x)) + 2 cdot frac(1)(cos(x)) cdot frac(sin(x))(cos(x)) + frac(sin^(2)(x))(cos^(2)(x))=1cos2(x)+2⋅1cos(x)⋅sin(x)cos(x)+sin2(x)cos2(x)
= frac(1 + sin^(2)(x))(cos^(2)(x)) + frac(2 sin(x))(cos^(2)(x))=1+sin2(x)cos2(x)+2sin(x)cos2(x)
= frac(sin^(2)(x) + 2 sin(x) + 1)(cos^(2)(x))=sin2(x)+2sin(x)+1cos2(x)
Let's factorise the numerator of the fraction using the "middle-term break":
= frac(sin^(2)(x) + sin(x) + sin(x) + 1)(cos^(2)(x))=sin2(x)+sin(x)+sin(x)+1cos2(x)
= frac(sin(x)(sin(x) + 1) + 1 (sin(x) + 1))(cos^(2)(x))=sin(x)(sin(x)+1)+1(sin(x)+1)cos2(x)
= frac((sin(x) + 1)(sin(x) + 1))(cos^(2)(x))=(sin(x)+1)(sin(x)+1)cos2(x)
= frac((sin(x) + 1)^(2))(cos^(2)(x))=(sin(x)+1)2cos2(x)
One of the Pythagorean identities is cos^(2)(x) + sin^(2)(x) = 1cos2(x)+sin2(x)=1.
We can rearrange it to get:
Rightarrow cos^(2)(x) = 1 - sin^(2)(x)⇒cos2(x)=1−sin2(x)
Let's apply this rearranged identity to our proof:
= frac((1 + sin(x))^(2))(1 - sin^(2)(x))=(1+sin(x))21−sin2(x)
Let's factorise the denominator, which is in the form of a difference of two squares:
= frac((1 + sin(x))^(2))((1 + sin(x))(1 - sin(x)))=(1+sin(x))2(1+sin(x))(1−sin(x))
= frac(1 + sin(x))(1 - sin(x)) " " " =1+sin(x)1−sin(x) "Q.E.D."Q.E.D.
