To determine whether the given trigonometric identity is true or false, let's analyze the expression:
[tex]\[ \tan^2 x = \frac{1 - \cos 2x}{1 + \cos 2x} \][/tex]
First, recall the double angle identity for cosine:
[tex]\[ \cos 2x = 2\cos^2 x - 1 \][/tex]
But we can also use:
[tex]\[ \cos 2x = 1 - 2\sin^2 x \][/tex]
Now, let’s use a known identity related to tangent:
[tex]\[ \tan x = \frac{\sin x}{\cos x} \][/tex]
[tex]\[ \tan^2 x = \left( \frac{\sin x}{\cos x} \right)^2 = \frac{\sin^2 x}{\cos^2 x} \][/tex]
Next, we apply the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \][/tex]
Thus:
[tex]\[ \sin^2 x = 1 - \cos^2 x \][/tex]
So:
[tex]\[ \tan^2 x = \frac{1 - \cos^2 x}{\cos^2 x} \][/tex]
Consider the double angle identities again:
[tex]\[ \cos 2x = 2\cos^2 x - 1 \][/tex]
[tex]\[ 1 - \cos 2x = 1 - (2\cos^2 x - 1) = 2 - 2\cos^2 x = 2(1 - \cos^2 x) = 2\sin^2 x \][/tex]
Similarly:
[tex]\[ 1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x \][/tex]
Thus:
[tex]\[ \frac{1 - \cos 2x}{1 + \cos 2x} = \frac{2\sin^2 x}{2\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x \][/tex]
Therefore, given the derivation:
[tex]\[ \tan^2 x = \frac{1 - \cos 2x}{1 + \cos 2x} \][/tex]
The provided trigonometric identity is true. So the correct answer is:
A. True
