In a `DeltaABC, tanA and tanB` are the roots of the equation `ab(x^(2)+1)=c^(2)x`, where a, b and c are the sides of the triangle. Then

Options

A. `tan(A-B)=(a^(2)-b^(2))/(2ab)`
B. `cotC=0`
C. `sin^(2)A+sin^(2)B=1`
D. none of these

Answer

To solve the problem, we start with the given equation:

\[ ab(x^2 + 1) = c^2 x \]

This can be rearranged into the standard form of a quadratic equation:

\[ ab x^2 - c^2 x + ab = 0 \]

Here, we recognize that the roots of this quadratic equation are \( \tan A \) and \( \tan B \).

### Step 1: Identify the coefficients
From the quadratic equation \( ax^2 + bx + c = 0 \), we can identify:
- \( a = ab \)
- \( b = -c^2 \)
- \( c = ab \)

### Step 2: Use the relationship of roots
The sum and product of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) are given by:
- Sum of roots \( \tan A + \tan B = -\frac{b}{a} \)
- Product of roots \( \tan A \tan B = \frac{c}{a} \)

Substituting the identified coefficients:
- Sum of roots:
\[ \tan A + \tan B = -\frac{-c^2}{ab} = \frac{c^2}{ab} \]
- Product of roots:
\[ \tan A \tan B = \frac{ab}{ab} = 1 \]

### Step 3: Use the tangent addition formula
From the tangent addition formula, we know:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Substituting the values we found:
\[ \tan(A + B) = \frac{\frac{c^2}{ab}}{1 - 1} = \text{undefined} \]
This indicates that \( A + B = \frac{\pi}{2} \).

### Step 4: Find angle C
Since \( A + B + C = \pi \) in triangle ABC, we can find angle C:
\[ C = \pi - (A + B) = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]

### Step 5: Use the Pythagorean theorem
In a right triangle (since \( C = \frac{\pi}{2} \)), we have:
\[ a^2 + b^2 = c^2 \]

### Conclusion
Thus, we conclude that in triangle ABC, if \( \tan A \) and \( \tan B \) are the roots of the given equation, then angle C must be \( \frac{\pi}{2} \) and the relationship between the sides is given by:
\[ a^2 + b^2 = c^2 \]