If `y=(tan x + cot x)/(tan x - cot x)`, then `(dy)/(dx)=`

Options

A. 2 tan 2x sec 2x
B. tan 2x sec 2x
C. `-tan 2x sec 2x`
D. `-2 tan 2x sec 2x`

Answer

To solve the problem \( y = \frac{\tan x + \cot x}{\tan x - \cot x} \) and find \( \frac{dy}{dx} \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the given expression:
\[ y = \frac{\tan x + \cot x}{\tan x - \cot x} \]

### Step 2: Multiply numerator and denominator by \(\tan x\)
To simplify, we multiply the numerator and denominator by \(\tan x\):
\[ y = \frac{\tan^2 x + 1}{\tan^2 x - 1} \]
Here, we used the identity \(\cot x = \frac{1}{\tan x}\).

### Step 3: Use the identity for \(\tan^2 x\)
Recall that \(\tan^2 x + 1 = \sec^2 x\) and \(\tan^2 x - 1 = \sec^2 x - 2\):
\[ y = \frac{\sec^2 x}{\sec^2 x - 2} \]

### Step 4: Rewrite the expression
We can rewrite \(y\) as:
\[ y = \frac{1}{\frac{\sec^2 x - 2}{\sec^2 x}} = \frac{1}{1 - \frac{2}{\sec^2 x}} = \frac{1}{1 - 2 \cos^2 x} \]

### Step 5: Differentiate using the quotient rule
To find \(\frac{dy}{dx}\), we will differentiate \(y\) using the quotient rule:
\[ \frac{dy}{dx} = \frac{0 \cdot (1 - 2 \cos^2 x) - 1 \cdot (-4 \cos x \sin x)}{(1 - 2 \cos^2 x)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{4 \cos x \sin x}{(1 - 2 \cos^2 x)^2} \]

### Step 6: Simplify further
Recognizing that \(2 \cos x \sin x = \sin 2x\), we can write:
\[ \frac{dy}{dx} = \frac{2 \sin 2x}{(1 - 2 \cos^2 x)^2} \]

### Final Result
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{2 \sin 2x}{(1 - 2 \cos^2 x)^2} \]