`sin^(-1)(sin((tan^(-1)1 +tan^(-1) 2 +tan^(-1)3)/(cot^(-1)1+cot^(-1)2+cot^(-1)3)))=`

Options

A. 0
B. `pi-2`
C. `2-pi`
D. 2

Answer

To solve the given problem step by step, we will use the properties of inverse trigonometric functions and the formulas for the sum of inverse tangents and cotangents.

### Step 1: Understand the Expression
We need to evaluate:
\[ \sin^{-1}\left(\sin\left(\frac{\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)}{\cot^{-1}(1) + \cot^{-1}(2) + \cot^{-1}(3)}\right)\right) \]

### Step 2: Use the Formula for Sum of Inverse Tangents
Recall the formula:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \text{ if } xy < 1 \]
We will apply this formula to simplify \(\tan^{-1}(1) + \tan^{-1}(2)\).

Let \(x = 1\) and \(y = 2\):
\[ \tan^{-1}(1) + \tan^{-1}(2) = \tan^{-1}\left(\frac{1 + 2}{1 - 1 \cdot 2}\right) = \tan^{-1}\left(\frac{3}{-1}\right) = \tan^{-1}(-3) \]

Now, we add \(\tan^{-1}(3)\):
\[ \tan^{-1}(-3) + \tan^{-1}(3) = \tan^{-1}\left(\frac{-3 + 3}{1 - (-3)(3)}\right) = \tan^{-1}\left(\frac{0}{1 + 9}\right) = \tan^{-1}(0) \]

### Step 3: Evaluate the Denominator
For the denominator, we have:
\[ \cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x) \]
Thus:
\[ \cot^{-1}(1) + \cot^{-1}(2) + \cot^{-1}(3) = \left(\frac{\pi}{2} - \tan^{-1}(1)\right) + \left(\frac{\pi}{2} - \tan^{-1}(2)\right) + \left(\frac{\pi}{2} - \tan^{-1}(3)\right) \]
This simplifies to:
\[ \frac{3\pi}{2} - \left(\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)\right) \]
Using the previous result:
\[ \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \tan^{-1}(0) = 0 \]
So:
\[ \cot^{-1}(1) + \cot^{-1}(2) + \cot^{-1}(3) = \frac{3\pi}{2} \]

### Step 4: Substitute Back into the Expression
Now we substitute back into the original expression:
\[ \frac{\tan^{-1}(0)}{\frac{3\pi}{2}} = \frac{0}{\frac{3\pi}{2}} = 0 \]

### Step 5: Final Evaluation
Now we evaluate:
\[ \sin^{-1}(\sin(0)) = \sin^{-1}(0) = 0 \]

### Conclusion
Thus, the final answer is:
\[ \boxed{0} \]