Verifying Trigonometric Identities: $2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$ Verify that $$ 2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$$

It's generally easier to turn more-complicated stuff into less-complicated stuff; and one common suggestion is always "first convert to sine and cosine":
$$\begin{align}
\frac{1-\tan^2\theta}{1+\tan^2\theta} 
&= \frac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}} \\
&=\frac{\left(\cos^2\theta - \sin^2\theta\right)/\cos^2\theta}{\left(\cos^2\theta + \sin^2\theta\right)/\cos^2\theta}\\
&=\frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta + \sin^2\theta} &\text{denom = 1}\\[6pt]
&= \cos^2\theta - \sin^2\theta &(1)\\
&= \cos^2\theta - \left( 1 - \cos^2\theta \right) \\
&= 2 \cos^2\theta - 1 &(2)
\end{align}$$
You can jump from (1) to (2) by recognizing both as forms of $\cos 2\theta$.
With a little more fluency in interrelations between more of the trig functions, you might also have taken this route:
$$\begin{align}
\frac{1-\tan^2\theta}{1+\tan^2\theta} &= \frac{1-\tan^2\theta}{\sec^2\theta} \\
&=\left( 1 - \tan^2\theta \right) \cos^2\theta \\
&=\cos^2\theta - \sin^2\theta \\[6pt]
&= 2\cos^2\theta - 1 
\end{align}$$  
