In a quadrilateral ABCD , `angleD=270^(@)` then `tan A . Tan B + tan B . Tan C+tanC.tanA=`?

Options

A. 0
B. 1
C. 3
D. Can't determined

Answer

To solve the problem, we need to find the value of \( \tan A \cdot \tan B + \tan B \cdot \tan C + \tan C \cdot \tan A \) given that \( \angle D = 270^\circ \) in quadrilateral \( ABCD \).

### Step-by-Step Solution:

1. **Understanding the Angles in Quadrilateral**:
The sum of the angles in any quadrilateral is \( 360^\circ \). Given that \( \angle D = 270^\circ \), we can find the sum of the other angles:
\[ \angle A + \angle B + \angle C = 360^\circ - 270^\circ = 90^\circ \]

2. **Using the Tangent Addition Formula**:
We know that if \( A + B + C = 90^\circ \), we can use the tangent addition formula:
\[ \tan(A + B + C) = \tan(90^\circ) \text{ which is undefined (or infinite)} \]
This implies that:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \text{ (since } \tan(90^\circ) \text{ is infinite)} \]

3. **Rearranging the Equation**:
From the above, we can derive:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]

4. **Finding the Required Expression**:
We need to find the value of:
\[ \tan A \cdot \tan B + \tan B \cdot \tan C + \tan C \cdot \tan A \]
We can use the identity derived from the tangent addition:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]
Let \( x = \tan A \), \( y = \tan B \), and \( z = \tan C \). Therefore:
\[ x + y + z = xyz \]

5. **Using the Identity**:
The expression we need can be rewritten as:
\[ xy + yz + zx = 1 \]
This is a known result when \( x + y + z = xyz \).

6. **Conclusion**:
Therefore, the value of \( \tan A \cdot \tan B + \tan B \cdot \tan C + \tan C \cdot \tan A \) is:
\[ 1 \]

### Final Answer:
\[ \tan A \cdot \tan B + \tan B \cdot \tan C + \tan C \cdot \tan A = 1 \]

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