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The problem you've provided is related to proving certain trigonometric identities in a triangle \( \triangle ABC \). Let's break it down and solve it step by step.

### Problem Statement:

Given a triangle \( ABC \), prove the following:

#### a) \( \sin C = \sin A \cos B + \sin B \cos A \)

This is the standard **sine rule** for the sum of two angles:

\[
\sin(A + B) = \sin A \cos B + \cos A \sin B
\]

Since in a triangle, \( A + B + C = 180^\circ \), we have \( C = 180^\circ - (A + B) \). Therefore:

\[
\sin C = \sin(180^\circ - (A + B)) = \sin(A + B)
\]

So, this identity directly follows from the angle sum formula for sine.

#### b) \( \frac{\sin C}{\cos A \cos B} = \tan A + \tan B \) for \( A, B \neq 90^\circ \)

Starting with the left-hand side:

\[
\frac{\sin C}{\cos A \cos B}
\]

Using the identity from part (a):

\[
\sin C = \sin A \cos B + \cos A \sin B
\]

Now divide both sides of the equation by \( \cos A \cos B \):

\[
\frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} = \frac{\sin A \cos B}{\cos A \cos B} + \frac{\cos A \sin B}{\cos A \cos B}
\]

Simplifying:

\[
= \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} = \tan A + \tan B
\]

Thus, the identity is proven.

#### c) \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \) for \( A, B, C \neq 90^\circ \)

We start from the identity:

\[
\tan(A + B + C) = \tan 180^\circ = 0
\]

Since \( A + B + C = 180^\circ \) in a triangle, the left-hand side of this equation becomes:

\[
\tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}
\]

But since \( \tan 180^\circ = 0 \), the numerator must be zero, giving:

\[
\tan A + \tan B + \tan C = \tan A \tan B \tan C
\]

Thus, this identity is proven as well.

---

Would you like further clarification on any of these steps or additional details?

Here are 5 related questions to explore further:
1. How do the sine and cosine rules apply in non-right triangles?
2. Can we derive other identities involving tangent and cotangent in a triangle?
3. What happens to the trigonometric identities when one of the angles is \( 90^\circ \)?
4. How would these identities change if the triangle was isosceles or equilateral?
5. How are these identities useful in solving non-right triangle problems in geometry?

**Tip**: When solving trigonometric identities, try to express all terms using basic identities like \( \sin(A + B) \), \( \cos(A + B) \), and \( \tan(A + B) \) to simplify proofs.

Given a triangle ABC, prove the following: a) sinC = sinA*cosB + sinB*cosA, b) (sinC / cosA*cosB) = tanA + tanB, c) tanA + tanB + tanC = tanA * tanB * tanC

Learn how to prove trigonometric identities involving sine and tangent functions in triangle ABC. This problem involves using the sine rule and tangent identity formulas. Suitable for students in Grades 10-12.

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