\( 4.2 \quad \frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } = 2 \cos ^ { 2 } \theta - 1 \)

To solve the equation \( 4.2 \quad \frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } = 2 \cos ^ { 2 } \theta - 1 \), we start by recognizing that the left side resembles the tangent double angle identity. We rewrite the left-hand side using the identity \( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta \). Thus, we have: \[ 4.2 \cos 2\theta = 2 \cos^2 \theta - 1 \] Now, note that the right side, \( 2 \cos^2 \theta - 1 \), is equal to \( \cos 2\theta \) as well. Thus, we can simplify our equation: \[ 4.2 \cos 2\theta = \cos 2\theta \] If we factor this out, we get: \[ \cos 2\theta (4.2 - 1) = 0 \] This implies: \[ \cos 2\theta = 0 \quad \text{or} \quad 4.2 - 1 = 3.2 = 0 \] Since \( 3.2 \neq 0 \), we focus on \( \cos 2\theta = 0 \). This gives us: \[ 2\theta = (2n + 1)\frac{\pi}{2}, n \in \mathbb{Z} \] Thus, solving for \(\theta\) yields: \[ \theta = (2n + 1)\frac{\pi}{4}, n \in \mathbb{Z} \] These are the angles where the original equation holds true!