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 can follow these steps:

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

### Step 2: Find \( \cos \theta \)
Using the Pythagorean identity:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Substituting \( \sin \theta = x \):
\[ \cos^2 \theta + x^2 = 1 \]
Thus,
\[ \cos^2 \theta = 1 - x^2 \]
Taking the square root:
\[ \cos \theta = \sqrt{1 - x^2} \quad \text{(since } \theta \text{ is in the range of } \sin^{-1} \text{, } \cos \theta \text{ is non-negative)} \]

### Step 3: Express \( \sin^{-1} x \) in terms of \( \cos^{-1} \)
From the definition of cosine inverse:
\[ \theta = \cos^{-1}(\cos \theta) = \cos^{-1}(\sqrt{1 - x^2}) \]
Thus, we have:
\[ \sin^{-1} x = \cos^{-1} \sqrt{1 - x^2} \]

### Step 4: Express \( \sin^{-1} x \) in terms of \( \tan^{-1} \)
Using the definition of tangent:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x}{\sqrt{1 - x^2}} \]
Thus,
\[ \theta = \tan^{-1} \left( \frac{x}{\sqrt{1 - x^2}} \right) \]
So we have:
\[ \sin^{-1} x = \tan^{-1} \left( \frac{x}{\sqrt{1 - x^2}} \right) \]

### Step 5: Express \( \sin^{-1} x \) in terms of \( \cot^{-1} \)
Using the definition of cotangent:
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{1 - x^2}}{x} \]
Thus,
\[ \theta = \cot^{-1} \left( \frac{\sqrt{1 - x^2}}{x} \right) \]
So we have:
\[ \sin^{-1} x = \cot^{-1} \left( \frac{\sqrt{1 - x^2}}{x} \right) \]

### Final Results
We can summarize the results as follows:
1. \( \sin^{-1} x = \cos^{-1} \sqrt{1 - x^2} \)
2. \( \sin^{-1} x = \tan^{-1} \left( \frac{x}{\sqrt{1 - x^2}} \right) \)
3. \( \sin^{-1} x = \cot^{-1} \left( \frac{\sqrt{1 - x^2}}{x} \right) \)