If ABC D is a cyclic quadrilateral, then find the value of `sinA+sinB-sinC-sinD`
If `A+B+C=(pi)/(2)`, then find the value of `tanA tanB+tanBtanC+tanC tanA`

Answer

To solve the given problems step by step, we will break down each part of the question.

### Problem 1: Finding the value of \( \sin A + \sin B - \sin C - \sin D \)

1. **Understand the properties of a cyclic quadrilateral**: In a cyclic quadrilateral, the sum of opposite angles is \( 180^\circ \). Therefore, we have:
\[ A + D = 180^\circ \quad \text{(1)} \]
\[ B + C = 180^\circ \quad \text{(2)} \]

2. **Express \( C \) and \( D \) in terms of \( A \) and \( B \)**:
From equation (1), we can express \( D \):
\[ D = 180^\circ - A \]
From equation (2), we can express \( C \):
\[ C = 180^\circ - B \]

3. **Substitute \( C \) and \( D \) into the expression**:
We need to evaluate:
\[ \sin A + \sin B - \sin C - \sin D \]
Substituting \( C \) and \( D \):
\[ = \sin A + \sin B - \sin(180^\circ - B) - \sin(180^\circ - A) \]

4. **Use the sine property**:
We know that \( \sin(180^\circ - \theta) = \sin \theta \). Therefore:
\[ = \sin A + \sin B - \sin B - \sin A \]

5. **Simplify the expression**:
\[ = \sin A - \sin A + \sin B - \sin B = 0 \]

### Conclusion for Problem 1:
The value of \( \sin A + \sin B - \sin C - \sin D \) is \( 0 \).

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### Problem 2: Finding the value of \( \tan A \tan B + \tan B \tan C + \tan C \tan A \)

1. **Given condition**: We know that:
\[ A + B + C = \frac{\pi}{2} \]

2. **Use the tangent addition formula**:
The formula for \( \tan(A + B + C) \) is:
\[ \tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)} \]

3. **Substituting \( A + B + C = \frac{\pi}{2} \)**:
Since \( \tan\left(\frac{\pi}{2}\right) \) is undefined (approaches infinity), we can set the denominator to zero:
\[ 1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A) = 0 \]
Thus,
\[ \tan A \tan B + \tan B \tan C + \tan C \tan A = 1 \]

### Conclusion for Problem 2:
The value of \( \tan A \tan B + \tan B \tan C + \tan C \tan A \) is \( 1 \).

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