$$ \frac{1+ an x}{1- an x}=\frac{1+\sin 2 x}{\cos 2 x} $$

Question

$$ \frac{1+	an x}{1-	an x}=\frac{1+\sin 2 x}{\cos 2 x} $$

UpStudy AI · Accepted Answer

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
$$
x\neq\frac{\pi}{4}+\frac{k\pi}{2},\quad k\in \mathbb{Z}.
$$
The equation $$ \frac{1+\tan x}{1-\tan x} = \frac{1+\sin 2x}{\cos 2x} $$ is an identity, meaning it holds true for all real numbers $$ x $$ except when $$ x = \frac{\pi}{4} + \frac{k\pi}{2} $$ for any integer $$ k $$.

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