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The image presents three parts of a trigonometric proof problem for triangle \( ABC \). Here is the breakdown of the tasks:

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**Bài 4: Cho tam giác ABC. Chứng minh:**

**a)**
\[ \sin C = \sin A \cdot \cos B + \sin B \cdot \cos A \]

**b)**
\[ \frac{\sin C}{\cos A \cdot \cos B} = \tan A + \tan B \quad (A, B \neq 90^\circ) \]

**c)**
\[ \tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C \quad (A, B, C \neq 90^\circ) \]

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### Solution Approach:

### a) Prove the identity:
We use the **sine addition formula** to show the given identity.

The sum of angles in a triangle is \( A + B + C = 180^\circ \), so we can express \( C \) as \( C = 180^\circ - (A + B) \). The sine of \( C \) is:
\[
\sin C = \sin(180^\circ - (A + B)) = \sin(A + B)
\]
Now, applying the sum of angles identity for sine:
\[
\sin(A + B) = \sin A \cos B + \cos A \sin B
\]
Thus, we have:
\[
\sin C = \sin A \cos B + \sin B \cos A
\]
This proves the first part.

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### b) Prove the second identity:
We start with the left-hand side:
\[
\frac{\sin C}{\cos A \cos B}
\]
From part (a), we know \( \sin C = \sin A \cos B + \sin B \cos A \), so substituting

Prove the following for triangle ABC: a) sinC = sinA * cosB + sinB * cosA, b) sinC / (cosA * cosB) = tanA + tanB (A, B ≠ 90°), c) tanA + tanB + tanC = tanA * tanB * tanC (A, B, C ≠ 90°).

Learn how to prove trigonometric identities for triangle ABC. This problem covers the sine addition formula, tangent identities, and the sum of angles in a triangle. Suitable for Grades 10-12.

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