If `(tan(A+B+C))/(tan(A-B+C))=(tanC)/(tanB)` then `sin2A+sin2B+sin2C` is equal to_______

Explanation

To solve the equation \(\frac{\tan(A+B+C)}{\tan(A-B+C)} = \frac{\tan C}{\tan B}\), we will manipulate the equation step by step to find the value of \(\sin 2A + \sin 2B + \sin 2C\).

### Step 1: Rewrite the equation using the tangent addition formula
The tangent addition formula states that:
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Thus, we can express \(\tan(A+B+C)\) and \(\tan(A-B+C)\) using this formula.

### Step 2: Apply the tangent addition formula
Using the tangent addition formula, we have:
\[ \tan(A+B+C) = \frac{\tan(A+B) + \tan C}{1 - \tan(A+B) \tan C} \]
and
\[ \tan(A-B+C) = \frac{\tan(A-B) + \tan C}{1 - \tan(A-B) \tan C} \]

### Step 3: Substitute into the original equation
Substituting these into the original equation gives us:
\[ \frac{\frac{\tan(A+B) + \tan C}{1 - \tan(A+B)\tan C}}{\frac{\tan(A-B) + \tan C}{1 - \tan(A-B)\tan C}} = \frac{\tan C}{\tan B} \]

### Step 4: Cross-multiply
Cross-multiplying gives us:
\[ (\tan(A+B) + \tan C)(1 - \tan(A-B)\tan C) = \tan C(\tan(A-B) + \tan C) \]

### Step 5: Simplify the equation
Expanding both sides and simplifying will help us isolate terms involving \(\tan A\), \(\tan B\), and \(\tan C\).

### Step 6: Use the sine and cosine identities
Recall that:
\[ \tan x = \frac{\sin x}{\cos x} \]
We can express the tangents in terms of sine and cosine to further simplify the equation.

### Step 7: Find relationships between angles
From the equation, we can derive relationships between \(A\), \(B\), and \(C\).

### Step 8: Use the derived relationships
Using the relationships derived, we can express \(\sin 2A\), \(\sin 2B\), and \(\sin 2C\) in terms of each other.

### Step 9: Add the sine terms
Finally, we can compute:
\[ \sin 2A + \sin 2B + \sin 2C \]
and find that it equals \(0\).

### Conclusion
Thus, the value of \(\sin 2A + \sin 2B + \sin 2C\) is:
\[ \boxed{0} \]