`(d)/(dx)[tan^(-1)(cotx)+cot^(-1)(tanx)]=`

Options

A. 0
B. 1
C. `-1`
D. `-2`

Answer

To solve the problem \(\frac{d}{dx}\left[\tan^{-1}(\cot x) + \cot^{-1}(\tan x)\right]\), we will differentiate each term separately using the chain rule and the derivatives of inverse trigonometric functions.

### Step 1: Differentiate \(\tan^{-1}(\cot x)\)

The derivative of \(\tan^{-1}(u)\) with respect to \(x\) is given by:
\[ \frac{d}{dx}[\tan^{-1}(u)] = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \(u = \cot x\).

Now, we need to find \(\frac{du}{dx}\):
\[ \frac{d}{dx}[\cot x] = -\csc^2 x \]

Now, substituting \(u = \cot x\):
\[ \frac{d}{dx}[\tan^{-1}(\cot x)] = \frac{1}{1 + \cot^2 x} \cdot (-\csc^2 x) \]

Using the identity \(1 + \cot^2 x = \csc^2 x\):
\[ \frac{d}{dx}[\tan^{-1}(\cot x)] = \frac{-\csc^2 x}{\csc^2 x} = -1 \]

### Step 2: Differentiate \(\cot^{-1}(\tan x)\)

The derivative of \(\cot^{-1}(u)\) with respect to \(x\) is given by:
\[ \frac{d}{dx}[\cot^{-1}(u)] = -\frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \(u = \tan x\).

Now, we need to find \(\frac{du}{dx}\):
\[ \frac{d}{dx}[\tan x] = \sec^2 x \]

Now, substituting \(u = \tan x\):
\[ \frac{d}{dx}[\cot^{-1}(\tan x)] = -\frac{1}{1 + \tan^2 x} \cdot \sec^2 x \]

Using the identity \(1 + \tan^2 x = \sec^2 x\):
\[ \frac{d}{dx}[\cot^{-1}(\tan x)] = -\frac{\sec^2 x}{\sec^2 x} = -1 \]

### Step 3: Combine the derivatives

Now, we can combine the derivatives from Step 1 and Step 2:
\[ \frac{d}{dx}\left[\tan^{-1}(\cot x) + \cot^{-1}(\tan x)\right] = -1 - 1 = -2 \]

### Final Answer
\[ \frac{d}{dx}\left[\tan^{-1}(\cot x) + \cot^{-1}(\tan x)\right] = -2 \]

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