Express `sin^(-1)x` in terms of (i) `cos^(-1)sqrt(1-x^(2))` (ii) `"tan"^(-1)x/(sqrt(1-x^(2)))` (iii) `"cot"^(-1)(sqrt(1-x^(2)))/x`

Explanation

To express \( \sin^{-1} x \) in terms of \( \cos^{-1} \sqrt{1-x^2} \), \( \tan^{-1} \frac{x}{\sqrt{1-x^2}} \), and \( \cot^{-1} \frac{\sqrt{1-x^2}}{x} \), we will follow these steps:

### Step 1: Define \( \sin^{-1} x \)
Let \( \theta = \sin^{-1} x \). This implies:
\[ \sin \theta = x \]

### Step 2: Find \( \cos \theta \)
Using the Pythagorean identity:
\[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - x^2} \]

### Step 3: Express \( \sin^{-1} x \) in terms of \( \cos^{-1} \)
From the relationship between sine and cosine:
\[ \cos \theta = \sqrt{1 - x^2} \implies \theta = \cos^{-1}(\sqrt{1 - x^2}) \]
Thus, we can write:
\[ \sin^{-1} x = \cos^{-1}(\sqrt{1 - x^2}) \quad \text{for } x \in [0, 1] \]
For \( x \in [-1, 0) \):
\[ \sin^{-1} x = -\cos^{-1}(\sqrt{1 - x^2}) \]

### Step 4: Find \( \tan \theta \)
Using the definition of tangent:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x}{\sqrt{1 - x^2}} \]
Thus, we can express \( \theta \) in terms of \( \tan^{-1} \):
\[ \sin^{-1} x = \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) \quad \text{for } x \in [-1, 1] \]

### Step 5: Find \( \cot \theta \)
Using the definition of cotangent:
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{1 - x^2}}{x} \]
Thus, we can express \( \theta \) in terms of \( \cot^{-1} \):
\[ \sin^{-1} x = \cot^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right) \quad \text{for } x \in (0, 1] \]
For \( x \in [-1, 0) \):
\[ \sin^{-1} x = \cot^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right) - \pi \]

### Final Expressions
Thus, we have:
1. \( \sin^{-1} x = \cos^{-1}(\sqrt{1 - x^2}) \) for \( x \in [0, 1] \) and \( \sin^{-1} x = -\cos^{-1}(\sqrt{1 - x^2}) \) for \( x \in [-1, 0) \).
2. \( \sin^{-1} x = \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) \) for \( x \in [-1, 1] \).
3. \( \sin^{-1} x = \cot^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right) \) for \( x \in (0, 1] \) and \( \sin^{-1} x = \cot^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right) - \pi \) for \( x \in [-1, 0) \).