prove that
(cosx - sinx) / (cosx + sinx) = sec2x - tan2x
multiply top and bottom by (cosx - sinx) to get
(cosx-sinx)^2 / (cos^2x - sin^2x)
(cos^2x - 2sinx cosx + sin^2x) / cos2x
(1 - sin2x)/cos2x
1/cos2x - sin2x/cos2x
QED
(cosx - sinx)^2 / (cosx + sinx)(cosx - sin x)
= (cos^2x - 2 sinx cosx + sin^2x) / (cos^2x - sin^2 x)
= (1 -2 sinx cosx) / cos2x
= 1/cos2x - sin2x /cos2x
= sec2x - tan2x
whew !
To prove the given equation:
(cosx - sinx) / (cosx + sinx) = sec2x - tan2x,
we can start with the left-hand side (LHS) and simplify it step by step until it matches the right-hand side (RHS).
Using the Difference of Squares identity (a^2 - b^2 = (a + b)(a - b)), we can rewrite the denominator:
(cosx + sinx) = (cosx + sinx)(cosx - sinx) / (cosx - sinx)
= (cos^2x - sin^2x) / (cosx - sinx)
Now, let's substitute the denominator in the LHS expression with the simplified form:
(cosx - sinx) / (cosx + sinx) = (cosx - sinx) * (cosx - sinx) / (cos^2x - sin^2x)
Expanding the numerator, we have:
= (cos^2x - 2sinxcosx + sin^2x) / (cos^2x - sin^2x)
Applying the Pythagorean identity (sin^2x + cos^2x = 1), we get:
= (cos^2x - 2sinxcosx + sin^2x) / (1 - sin^2x)
Rearranging the terms in the numerator and denominator, we have:
= (cos^2x + sin^2x - 2sinxcosx) / (1 - sin^2x)
Now, apply the trigonometric identity for double angle:
= (1 - 2sinxcosx) / (1 - sin^2x)
Next, apply the identity for 2sinθcosθ = sin(2θ):
= (1 - sin(2x)) / (1 - sin^2x)
Finally, use the Pythagorean identity again and simplify:
= (1 - sin(2x)) / cos^2x
= sec^2x - sin(2x) / cos^2x
Since sin(2x) = 2sinxcosx, we can substitute this into the equation:
= sec^2x - (2sinxcosx / cos^2x)
= sec^2x - 2tanx
= sec^2x - tan^2x
Therefore, we have proven that (cosx - sinx) / (cosx + sinx) is equal to sec^2x - tan^2x.