If `y=sec^(-1)((x+1)/(x-1))+sin^(-1)((x-1)/(x+1))` then `(dy)/(dx)` is equal to :

Options

A. A) 0
B. B) 1
C. C) `(x-1)/(x+1)`
D. D) `(x+1)/(x-1)`

Answer

To solve the problem, we start with the given function:

\[ y = \sec^{-1}\left(\frac{x+1}{x-1}\right) + \sin^{-1}\left(\frac{x-1}{x+1}\right) \]

### Step 1: Rewrite the Inverse Functions
We can use the property of inverse trigonometric functions. Recall that:

\[ \sec^{-1}(a) + \sin^{-1}(b) = \frac{\pi}{2} \text{ if } a = \frac{1}{b} \]

In our case, we need to check if:

\[ \frac{x+1}{x-1} = \frac{1}{\frac{x-1}{x+1}} \]

This simplifies to:

\[ \frac{x+1}{x-1} = \frac{x+1}{x-1} \]

This is true, so we can use the property:

\[ \sec^{-1}\left(\frac{x+1}{x-1}\right) + \sin^{-1}\left(\frac{x-1}{x+1}\right) = \frac{\pi}{2} \]

### Step 2: Simplify the Expression for y
Thus, we can rewrite y as:

\[ y = \frac{\pi}{2} \]

### Step 3: Differentiate y with Respect to x
Now, we differentiate y with respect to x:

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) \]

Since \(\frac{\pi}{2}\) is a constant, its derivative is:

\[ \frac{dy}{dx} = 0 \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is equal to:

\[ \frac{dy}{dx} = 0 \]