Let the slope of the curve `y = cos^(-1)` (sin x) be tan `theta` : Then the value of `theta` in the interval `(0, pi)` is

Options

A. `(pi)/(6)`
B. `(3pi)/(4)`
C. `(pi)/(4)`
D. `(pi)/(2)`

Explanation

To solve the problem, we need to find the slope of the curve given by \( y = \cos^{-1}(\sin x) \) and relate it to \( \tan \theta \).

### Step 1: Differentiate \( y = \cos^{-1}(\sin x) \)

To find the slope of the curve, we need to differentiate \( y \) with respect to \( x \).

Using the chain rule:
\[ \frac{dy}{dx} = \frac{d}{dx}(\cos^{-1}(u)) \cdot \frac{du}{dx} \]
where \( u = \sin x \).

The derivative of \( \cos^{-1}(u) \) is:
\[ \frac{d}{du}(\cos^{-1}(u)) = -\frac{1}{\sqrt{1 - u^2}} \]
Thus,
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - \sin^2 x}} \cdot \frac{d}{dx}(\sin x) \]

The derivative of \( \sin x \) is \( \cos x \), so:
\[ \frac{dy}{dx} = -\frac{\cos x}{\sqrt{1 - \sin^2 x}} \]

### Step 2: Simplify \( \sqrt{1 - \sin^2 x} \)

Using the identity \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sqrt{1 - \sin^2 x} = \sqrt{\cos^2 x} = |\cos x| \]

Since \( x \) is in the interval \( (0, \pi) \), \( \cos x \) is non-negative in \( (0, \frac{\pi}{2}) \) and non-positive in \( (\frac{\pi}{2}, \pi) \). Therefore, we can write:
\[ \sqrt{1 - \sin^2 x} = \cos x \quad \text{for } x \in (0, \frac{\pi}{2}) \]
and
\[ \sqrt{1 - \sin^2 x} = -\cos x \quad \text{for } x \in (\frac{\pi}{2}, \pi) \]

### Step 3: Substitute back into the derivative

Thus, for \( x \in (0, \frac{\pi}{2}) \):
\[ \frac{dy}{dx} = -\frac{\cos x}{\cos x} = -1 \]

For \( x \in (\frac{\pi}{2}, \pi) \):
\[ \frac{dy}{dx} = -\frac{\cos x}{-\cos x} = 1 \]

### Step 4: Relate to \( \tan \theta \)

We are given that the slope \( \frac{dy}{dx} \) is equal to \( \tan \theta \). Therefore, we have two cases:

1. For \( x \in (0, \frac{\pi}{2}) \):
\[ \tan \theta = -1 \implies \theta = \frac{3\pi}{4} \quad (\text{since } \theta \text{ must be in } (0, \pi)) \]

2. For \( x \in (\frac{\pi}{2}, \pi) \):
\[ \tan \theta = 1 \implies \theta = \frac{\pi}{4} \quad (\text{not valid since } \tan \theta = 1 \text{ does not yield a negative slope}) \]

### Conclusion

Thus, the only valid solution for \( \theta \) in the interval \( (0, \pi) \) is:
\[ \theta = \frac{3\pi}{4} \]