How do you prove #(1 - tanx) / (1 + tanx) = (1 - sin2x) / (cos2x)#?

Question

Zoe Austin · Accepted Answer

To find the solution, multiply the left side of the denominator and numerator by 1-tanx and then simplify: #(1-tanx)^2 / (1-tan^2 x)# =#(1+tan^2x -2tanx)/(1- (sin^2x/cos^2x)#    = #(sec^2x -2tanx)/((cos^2x -sin^2x)/cos^2x)# (because #1+tan^2x = sec^2x#) =#(1-2tan x cos^2x)/cos(2x)#  (because #cos^2x- sin^2x=cos 2x#) =#(1-sin2x)/cos(2x)#

Grace Ashcraft · Answer

undefined To prove $ \frac{1 - \tan(x)}{1 + \tan(x)} = \frac{1 - \sin(2x)}{\cos(2x)} $, we can start with the left side of the equation and manipulate it using trigonometric identities.

Starting with the left side:

$$ \frac{1 - \tan(x)}{1 + \tan(x)} $$

We can rewrite $ \tan(x) $ in terms of sine and cosine using the identity $ \tan(x) = \frac{\sin(x)}{\cos(x)} $:

$$ = \frac{1 - \frac{\sin(x)}{\cos(x)}}{1 + \frac{\sin(x)}{\cos(x)}} $$

Now, to simplify this expression, we can find a common denominator:

$$ = \frac{\cos(x) - \sin(x)}{\cos(x) + \sin(x)} $$

Next, we can multiply both the numerator and denominator by $ \cos(x) $ to eliminate the fraction:

$$ = \frac{\cos(x)(\cos(x) - \sin(x))}{\cos(x)(\cos(x) + \sin(x))} $$

Expanding the numerator and denominator:

$$ = \frac{\cos^2(x) - \cos(x)\sin(x)}{\cos^2(x) + \cos(x)\sin(x)} $$

Using the double-angle identity for sine, $ \sin(2x) = 2\sin(x)\cos(x) $, and the Pythagorean identity for cosine, $ \cos^2(x) = 1 - \sin^2(x) $:

$$ = \frac{1 - \sin^2(x) - \sin(x)\cos(x)}{1 - \sin^2(x) + \sin(x)\cos(x)} $$

Now, factor out $ 1 - \sin^2(x) $ from both the numerator and the denominator:

$$ = \frac{(1 - \sin^2(x))(1 - \sin(x))}{(1 - \sin^2(x))(1 + \sin(x))} $$

Cancel out the common factor of $ 1 - \sin^2(x) $:

$$ = \frac{1 - \sin(x)}{1 + \sin(x)} $$

Finally, using the double-angle identity for sine, $ \sin(2x) = 2\sin(x)\cos(x) $, and the Pythagorean identity for cosine, $ \cos(2x) = 1 - 2\sin^2(x) $:

$$ = \frac{1 - \sin(2x)}{\cos(2x)} $$

This matches the right side of the equation, proving the identity.