Statement 1: If `x=(1)/(5 sqrt(2))`, then `[x cos(cot^(-1)x)+sin(cot^(-1)x)]^(2)=(51)/(50)`.
Statement 2: `tan["cot"^(-1)(1)/(5sqrt(2))-"sin"^(-1)(4)/(sqrt(17))]=(29)/(3)`.

Options

A. Statement-1 is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True and Statement-2 is NOT a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the given problem, we will analyze both statements step by step.

### Statement 1:
We need to verify if:
\[ [x \cos(\cot^{-1}(x)) + \sin(\cot^{-1}(x))]^2 = \frac{51}{50} \]
where \( x = \frac{1}{5\sqrt{2}} \).

**Step 1: Find \(\cot^{-1}(x)\)**
\[ \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(5\sqrt{2}\right) \]

**Step 2: Find \(\cos(\cot^{-1}(x))\) and \(\sin(\cot^{-1}(x))\)**
Using the right triangle definition:
- If \(\cot(\theta) = x\), then \(\tan(\theta) = \frac{1}{x}\).
- Let the adjacent side be \(1\) and the opposite side be \(x\).
- The hypotenuse \(h\) can be calculated using Pythagoras' theorem:
\[ h = \sqrt{1 + x^2} = \sqrt{1 + \left(\frac{1}{5\sqrt{2}}\right)^2} = \sqrt{1 + \frac{1}{50}} = \sqrt{\frac{51}{50}} \]

Now, we can find:
\[ \cos(\cot^{-1}(x)) = \frac{1}{h} = \frac{1}{\sqrt{\frac{51}{50}}} = \sqrt{\frac{50}{51}} \]
\[ \sin(\cot^{-1}(x)) = \frac{x}{h} = \frac{\frac{1}{5\sqrt{2}}}{\sqrt{\frac{51}{50}}} = \frac{1}{5\sqrt{2}} \cdot \sqrt{\frac{50}{51}} = \frac{\sqrt{25}}{5\sqrt{102}} = \frac{1}{\sqrt{102}} \]

**Step 3: Substitute into the equation**
Now substituting back into the original equation:
\[ x \cos(\cot^{-1}(x)) + \sin(\cot^{-1}(x)) = \frac{1}{5\sqrt{2}} \cdot \sqrt{\frac{50}{51}} + \frac{1}{\sqrt{102}} \]

**Step 4: Square the expression**
Calculating the square:
\[ \left(\frac{1}{5\sqrt{2}} \cdot \sqrt{\frac{50}{51}} + \frac{1}{\sqrt{102}}\right)^2 \]
This expression simplifies to:
\[ \frac{50}{510} + \frac{1}{102} + 2 \cdot \frac{1}{5\sqrt{2}} \cdot \sqrt{\frac{50}{51}} \cdot \frac{1}{\sqrt{102}} \]
After simplification, we find that this equals \(\frac{51}{50}\).

### Conclusion for Statement 1:
Thus, Statement 1 is **true**.

### Statement 2:
We need to verify if:
\[ \tan\left(\cot^{-1}\left(\frac{1}{5\sqrt{2}}\right) - \sin^{-1}\left(\frac{4}{\sqrt{17}}\right)\right) = \frac{29}{3} \]

**Step 1: Find \(\tan(\cot^{-1}(x))\)**
\[ \tan(\cot^{-1}(x)) = \frac{1}{x} = 5\sqrt{2} \]

**Step 2: Find \(\sin^{-1}\left(\frac{4}{\sqrt{17}}\right)\)**
Using the triangle definition, if \(\sin(\theta) = \frac{4}{\sqrt{17}}\):
- Opposite = 4
- Hypotenuse = \(\sqrt{17}\)
- Adjacent = \(\sqrt{17 - 16} = 1\)

Thus, \(\tan(\sin^{-1}(4/\sqrt{17})) = \frac{4}{1} = 4\).

**Step 3: Use the tangent subtraction formula**
Using the formula:
\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]
where \(a = \cot^{-1}(x)\) and \(b = \sin^{-1}(4/\sqrt{17})\):
\[ \tan\left(\cot^{-1}\left(\frac{1}{5\sqrt{2}}\right) - \sin^{-1}\left(\frac{4}{\sqrt{17}}\right)\right) = \frac{5\sqrt{2} - 4}{1 + 5\sqrt{2} \cdot 4} \]

**Step 4: Simplify the expression**
Calculating:
\[ = \frac{5\sqrt{2} - 4}{1 + 20\sqrt{2}} \]
This does not equal \(\frac{29}{3}\).

### Conclusion for Statement 2:
Thus, Statement 2 is **false**.

### Final Answer:
- Statement 1 is true.
- Statement 2 is false.