Prove that 1-tan^2x/1+tan^2 =cos 2x

Answer 1

To prove this identity, we will start with the left-hand side (LHS) and use trigonometric identities to simplify it to the right-hand side (RHS).

LHS: (1 - tan^2x) / (1 + tan^2x)

First, we can express tan^2x in terms of sin^2x and cos^2x using the fundamental trigonometric identity:
tan^2x = sin^2x / cos^2x

Substituting this into the LHS, we get:
(1 - sin^2x / cos^2x) / (1 + sin^2x / cos^2x)

We can simplify the numerator by multiplying it by cos^2x:
cos^2x - sin^2x / cos^2x

Using the difference of squares formula (a^2 - b^2 = (a + b)(a - b)), we can further simplify the numerator:
[(cosx + sinx)(cosx - sinx)] / cos^2x

Now, we can simplify the denominator by multiplying it by cos^2x:
cos^2x + sin^2x / cos^2x

Again, using the identity sin^2x + cos^2x = 1, we can simplify the denominator to 1:
1 / cos^2x

Now, substituting these simplified expressions back into the LHS, we have:
[(cosx + sinx)(cosx - sinx)] / cos^2x / 1

Simplifying further, we get:
(cosx + sinx)(cosx - sinx) / cos^2x

Using the identity cos2x = cos^2x - sin^2x, we can rewrite the numerator:
(cosx + sinx)(cosx - sinx) = cos2x

Therefore, the LHS simplifies to the RHS, proving the identity:
LHS = RHS
(1 - tan^2x) / (1 + tan^2x) = cos2x