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

The equation you've provided is a trigonometric identity that equates two different expressions involving the angle \(\theta\): \[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = 2 \cos^2 \theta - 1 \] Let's verify and understand why both sides of this equation are equal. ### **Left Side: \(\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\)** 1. **Express \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\):** \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] 2. **Substitute \(\tan \theta\) into the expression:** \[ \frac{1 - \left(\frac{\sin \theta}{\cos \theta}\right)^2}{1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2} = \frac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 + \frac{\sin^2 \theta}{\cos^2 \theta}} \] 3. **Simplify the numerator and the denominator:** \[ \frac{\frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}}{\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}} = \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta + \sin^2 \theta} \] 4. **Recall the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\):** \[ \frac{\cos^2 \theta - \sin^2 \theta}{1} = \cos 2\theta \] So, \[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta \] ### **Right Side: \(2 \cos^2 \theta - 1\)** 1. **Use the double-angle identity for cosine:** \[ \cos 2\theta = 2 \cos^2 \theta - 1 \] So, \[ 2 \cos^2 \theta - 1 = \cos 2\theta \] ### **Conclusion** Both the left side and the right side of the equation simplify to \(\cos 2\theta\). Therefore, the original equation is a valid trigonometric identity: \[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = 2 \cos^2 \theta - 1 = \cos 2\theta \] This identity is useful in various trigonometric simplifications and transformations, especially when dealing with double-angle formulas.