If `tan^(-1)a+tan^(-1)b+tan^(-1)c=pi` then prove tjhat `a+b+c=abc`

Options

A. a+b+c=abc
B. ab+bc+ca=abc
C. `(1)/(a)+(1)/(b)+(1)/(c )-(1)/(abc)=0`
D. `ab+bc+ca=a+b+c`

Answer

To prove that if \( \tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi \), then \( a + b + c = abc \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ \tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi \]

### Step 2: Use the property of the tangent function
Using the property of the tangent function, we can express the sum of two inverse tangents:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
This holds true as long as \( ab < 1 \).

### Step 3: Rewrite the equation
We can rewrite the equation as:
\[ \tan^{-1} \left( \frac{a + b}{1 - ab} \right) + \tan^{-1} c = \pi \]

### Step 4: Apply the tangent function
Taking the tangent of both sides:
\[ \tan \left( \tan^{-1} \left( \frac{a + b}{1 - ab} \right) + \tan^{-1} c \right) = \tan(\pi) \]
Since \( \tan(\pi) = 0 \), we have:
\[ \tan \left( \tan^{-1} \left( \frac{a + b}{1 - ab} \right) + \tan^{-1} c \right) = 0 \]

### Step 5: Use the tangent addition formula
Using the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
we can express it as:
\[ \frac{\frac{a + b}{1 - ab} + c}{1 - \frac{a + b}{1 - ab} \cdot c} = 0 \]

### Step 6: Set the numerator to zero
For the fraction to equal zero, the numerator must be zero:
\[ \frac{a + b + c(1 - ab)}{1 - ab - c(a + b)} = 0 \]
Thus, we have:
\[ a + b + c(1 - ab) = 0 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ a + b + c - abc = 0 \]
or
\[ a + b + c = abc \]

### Conclusion
Thus, we have proved that if \( \tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi \), then:
\[ a + b + c = abc \]