Chapter 6 - Section 6.1 - Solving Trigonometric Equations - 6.1 Problem Set - Page 327: 83

$$\cos 2x = \frac{1-\tan^2x }{1+\tan^2x}$$

Since \begin{align*} \cos 2x&=\cos^2x-\sin^2x\\ &=\cos^2x\left(1-\frac{\sin^2x}{\cos^2x}\right)\\ &=\cos^2x\left(1-\tan^2x \right)\\ &=\frac{1-\tan^2x }{\sec^2x}\\ &= \frac{1-\tan^2x }{1+\tan^2x} \end{align*} Then $$\cos 2x = \frac{1-\tan^2x }{1+\tan^2x}$$