If `y=tan(x+y)`, then `(dy)/(dx) ` is equal to :

Options

A. `-"cot"^2(x+y)`
B. `cosec^2(x+y)`
C. cot (x+y)cosec (x+y)
D. `-tan^2 (x+y)`

Answer

To find \(\frac{dy}{dx}\) for the equation \(y = \tan(x + y)\), we will use implicit differentiation. Here are the steps:

### Step 1: Differentiate both sides with respect to \(x\)
We start with the equation:
\[ y = \tan(x + y) \]
Differentiating both sides with respect to \(x\):
\[ \frac{dy}{dx} = \sec^2(x + y) \cdot \left(1 + \frac{dy}{dx}\right) \]

### Step 2: Apply the chain rule
On the right side, we applied the chain rule. The derivative of \(\tan(u)\) is \(\sec^2(u) \cdot \frac{du}{dx}\), where \(u = x + y\). Thus:
\[ \frac{dy}{dx} = \sec^2(x + y) \cdot (1 + \frac{dy}{dx}) \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \sec^2(x + y) + \sec^2(x + y) \cdot \frac{dy}{dx} \]
Bringing all terms involving \(\frac{dy}{dx}\) to one side:
\[ \frac{dy}{dx} - \sec^2(x + y) \cdot \frac{dy}{dx} = \sec^2(x + y) \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} (1 - \sec^2(x + y)) = \sec^2(x + y) \]

### Step 4: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\sec^2(x + y)}{1 - \sec^2(x + y)} \]

### Step 5: Simplify the expression
Using the identity \(\sec^2(u) = 1 + \tan^2(u)\), we can rewrite:
\[ \frac{dy}{dx} = \frac{1 + \tan^2(x + y)}{1 - (1 + \tan^2(x + y))} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1 + \tan^2(x + y)}{-\tan^2(x + y)} = -\frac{1 + \tan^2(x + y)}{\tan^2(x + y)} \]

### Final Result
Thus, the final result for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -\frac{1 + y^2}{y^2} \]