If `x cancel(=) n pi ` where n is any integer and `( 1+ tan x )/ ( 1- tanx ) =1 + sin 2x`, then `tan x =`

Options

A. 1
B. `-1`
C. 3
D. 4

Answer

To solve the equation \( \frac{1 + \tan x}{1 - \tan x} = 1 + \sin 2x \) under the condition that \( x \neq n\pi \) (where \( n \) is any integer), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{1 + \tan x}{1 - \tan x} = 1 + \sin 2x \]
Recall that \( \sin 2x = 2 \sin x \cos x \). Therefore, we can rewrite the right-hand side:
\[ \frac{1 + \tan x}{1 - \tan x} = 1 + 2 \sin x \cos x \]

### Step 2: Express \( \tan x \) in terms of \( \sin x \) and \( \cos x \)
We know that \( \tan x = \frac{\sin x}{\cos x} \). Substitute this into the left-hand side:
\[ \frac{1 + \frac{\sin x}{\cos x}}{1 - \frac{\sin x}{\cos x}} = \frac{\cos x + \sin x}{\cos x - \sin x} \]

### Step 3: Set up the equation
Now we have:
\[ \frac{\cos x + \sin x}{\cos x - \sin x} = 1 + 2 \sin x \cos x \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ \cos x + \sin x = (1 + 2 \sin x \cos x)(\cos x - \sin x) \]

### Step 5: Expand the right-hand side
Expanding the right-hand side:
\[ \cos x + \sin x = \cos x - \sin x + 2 \sin x \cos^2 x - 2 \sin^2 x \cos x \]

### Step 6: Rearranging the equation
Rearranging terms leads to:
\[ \cos x + \sin x - \cos x + \sin x = 2 \sin x \cos^2 x - 2 \sin^2 x \cos x \]
This simplifies to:
\[ 2 \sin x = 2 \sin x \cos^2 x - 2 \sin^2 x \cos x \]

### Step 7: Factor out common terms
Factoring out \( 2 \sin x \) from both sides gives:
\[ 2 \sin x (1 - \cos^2 x + \sin x \cos x) = 0 \]
This implies either \( \sin x = 0 \) or \( 1 - \cos^2 x + \sin x \cos x = 0 \).

### Step 8: Solve for \( \tan x \)
1. If \( \sin x = 0 \), then \( x = n\pi \) (which we discard since \( x \neq n\pi \)).
2. For the second equation, we can use \( \sin^2 x + \cos^2 x = 1 \) to find \( \tan x \).

### Step 9: Find \( \tan x \)
From the equation \( \sin x + \cos x = 0 \), we can divide by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ \tan x = -1 \]

Thus, the final solution is:
\[ \tan x = -1 \]