For the curve defined as `y=cosec^(-1)((1+x^2)/(2x))+sec^(-1)((1+x^2)/(1-x^2))` the set of points at which the tangent is parallel to the x axis

Options

A. [0,1]
B. (0,1)
C. `(-1,0) cup (1,oo)`
D. None of these

Answer

To find the set of points at which the tangent to the curve defined by

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

is parallel to the x-axis, we need to determine where the derivative \( \frac{dy}{dx} \) is equal to zero.

### Step 1: Understanding the Function

The function \( y \) is defined as a sum of two inverse trigonometric functions. We will analyze each part separately.

### Step 2: Simplifying the Inverse Functions

1. **Cosecant Inverse**:
\[ \csc^{-1}(z) = \sin^{-1}\left(\frac{1}{z}\right) \]
So, we can rewrite:
\[ \csc^{-1}\left(\frac{1+x^2}{2x}\right) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \]

2. **Secant Inverse**:
\[ \sec^{-1}(z) = \cos^{-1}\left(\frac{1}{z}\right) \]
Thus, we can rewrite:
\[ \sec^{-1}\left(\frac{1+x^2}{1-x^2}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \]

### Step 3: Using Known Identities

Using the known identity:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
we can combine the two parts.

### Step 4: Combining the Results

Using the above identities, we can express \( y \):
\[ y = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \]

### Step 5: Finding the Derivative

Now, we need to find the derivative \( \frac{dy}{dx} \).

1. The derivative of a constant is zero.
2. If we can show that \( y \) simplifies to a constant, then \( \frac{dy}{dx} = 0 \).

### Step 6: Evaluating the Expression

After simplification, we find that:
\[ y = \pi \]
This indicates that \( y \) is a constant function.

### Step 7: Conclusion

Since \( y \) is constant, the derivative \( \frac{dy}{dx} = 0 \) for all values of \( x \). Therefore, the tangent to the curve is parallel to the x-axis at every point on the curve.

### Final Answer

The set of points at which the tangent is parallel to the x-axis is all points on the curve.

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