\( \frac{1+\tan x}{1-\tan x}=\frac{1+\sin 2 x}{\cos 2 x} \)

We start with the equation \[ \frac{1+\tan x}{1-\tan x}=\frac{1+\sin 2x}{\cos 2x}. \] **Step 1. Express the left‐hand side as a tangent of a sum** Recall that \[ \tan\left(\frac{\pi}{4}+x\right)=\frac{1+\tan x}{1-\tan x}. \] Thus the left side becomes \[ \tan\left(\frac{\pi}{4}+x\right). \] **Step 2. Rewrite the right‐hand side in terms of \(\tan x\) using double-angle formulas** Let \[ t=\tan x. \] We know the double-angle formulas: \[ \sin2x=\frac{2t}{1+t^2}\quad\text{and}\quad \cos2x=\frac{1-t^2}{1+t^2}. \] Substitute these into the right-hand side: \[ \frac{1+\sin2x}{\cos2x} = \frac{1+\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}} = \frac{\frac{1+t^2+2t}{1+t^2}}{\frac{1-t^2}{1+t^2}} = \frac{1+t^2+2t}{1-t^2}. \] Notice that the numerator is \[ 1+t^2+2t=(t+1)^2, \] and the denominator factors as \[ 1-t^2=(1-t)(1+t). \] Therefore, \[ \frac{1+\sin2x}{\cos2x}=\frac{(t+1)^2}{(1-t)(1+t)}=\frac{t+1}{1-t}, \] which is exactly the same as the expression for the left-hand side: \[ \frac{1+\tan x}{1-\tan x}=\frac{t+1}{1-t}. \] **Step 3. Conclude the equation is an identity** Since the left-hand side and the right-hand side simplify to the same expression (when defined), the equation is true for all \( x \) for which these expressions are defined. **Step 4. Determine the restrictions on \( x \)** The expressions are not defined when any denominator is zero. We have two restrictions: 1. From the left-hand side: \[ 1-\tan x \neq 0 \quad\Rightarrow\quad \tan x \neq 1. \] This implies \[ x \neq \frac{\pi}{4}+k\pi,\quad k\in \mathbb{Z}. \] 2. From the right-hand side: \[ \cos2x \neq 0. \] This means \[ 2x\neq \frac{\pi}{2}+k\pi,\quad k\in \mathbb{Z}\quad \Rightarrow\quad x\neq\frac{\pi}{4}+\frac{k\pi}{2},\quad k\in \mathbb{Z}. \] Thus, the equation holds for all \( x \) except when \[ x=\frac{\pi}{4}+\frac{k\pi}{2},\quad k\in \mathbb{Z}. \] **Final Answer** The given equation holds for all real \( x \) with the restriction \[ x\neq\frac{\pi}{4}+\frac{k\pi}{2},\quad k\in \mathbb{Z}. \]