`sin^-1(cos x)+tan ^-1(cot x)`

Answer

To solve the expression \( \sin^{-1}(\cos x) + \tan^{-1}(\cot x) \), we will differentiate it step by step.

### Step 1: Rewrite the expression
Let \( y = \sin^{-1}(\cos x) + \tan^{-1}(\cot x) \).

### Step 2: Differentiate \( y \)
We will differentiate \( y \) with respect to \( x \).

Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{d}{dx}(\sin^{-1}(\cos x)) + \frac{d}{dx}(\tan^{-1}(\cot x)) \]

### Step 3: Differentiate \( \sin^{-1}(\cos x) \)
Using the derivative of \( \sin^{-1}(u) \) which is \( \frac{1}{\sqrt{1 - u^2}} \) and applying the chain rule:
\[ \frac{d}{dx}(\sin^{-1}(\cos x)) = \frac{1}{\sqrt{1 - \cos^2 x}} \cdot (-\sin x) \]
Since \( 1 - \cos^2 x = \sin^2 x \), we can simplify this to:
\[ \frac{d}{dx}(\sin^{-1}(\cos x)) = \frac{-\sin x}{\sin x} = -1 \]

### Step 4: Differentiate \( \tan^{-1}(\cot x) \)
Using the derivative of \( \tan^{-1}(u) \) which is \( \frac{1}{1 + u^2} \) and applying the chain rule:
\[ \frac{d}{dx}(\tan^{-1}(\cot x)) = \frac{1}{1 + \cot^2 x} \cdot (-\csc^2 x) \]
Since \( 1 + \cot^2 x = \csc^2 x \), we can simplify this to:
\[ \frac{d}{dx}(\tan^{-1}(\cot x)) = \frac{-\csc^2 x}{\csc^2 x} = -1 \]

### Step 5: Combine the derivatives
Now, we combine the derivatives:
\[ \frac{dy}{dx} = -1 - 1 = -2 \]

### Final Answer
Thus, the derivative of \( y = \sin^{-1}(\cos x) + \tan^{-1}(\cot x) \) is:
\[ \frac{dy}{dx} = -2 \]

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