Let `(tan((pi)/(4)+alpha))/(5)=(tan((pi)/(4)+beta))/(3)=(tan((pi)/(4)+gamma))/(2)` . Then `12sin^(2)(alpha-beta)+15sin^(2)(beta-gamma)-7sin^(2)(gamma-alpha)` is equal to :

Options

A. `-(1)/(2)`
B. `(1)/(2)`
C. `1`
D. `0`

Answer

To solve the problem, we start with the equation given:

\[ \frac{\tan\left(\frac{\pi}{4} + \alpha\right)}{5} = \frac{\tan\left(\frac{\pi}{4} + \beta\right)}{3} = \frac{\tan\left(\frac{\pi}{4} + \gamma\right)}{2} \]

Let us denote this common value as \( k \). Thus, we can write:

\[ \tan\left(\frac{\pi}{4} + \alpha\right) = 5k \]
\[ \tan\left(\frac{\pi}{4} + \beta\right) = 3k \]
\[ \tan\left(\frac{\pi}{4} + \gamma\right) = 2k \]

Using the tangent addition formula, we have:

\[ \tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x} \]

Applying this to our equations:

1. For \( \tan\left(\frac{\pi}{4} + \alpha\right) = 5k \):
\[ 5k = \frac{1 + \tan \alpha}{1 - \tan \alpha} \]

2. For \( \tan\left(\frac{\pi}{4} + \beta\right) = 3k \):
\[ 3k = \frac{1 + \tan \beta}{1 - \tan \beta} \]

3. For \( \tan\left(\frac{\pi}{4} + \gamma\right) = 2k \):
\[ 2k = \frac{1 + \tan \gamma}{1 - \tan \gamma} \]

Next, we can solve for \( \tan \alpha \), \( \tan \beta \), and \( \tan \gamma \) in terms of \( k \):

From the first equation:
\[ 5k(1 - \tan \alpha) = 1 + \tan \alpha \implies 5k - 5k \tan \alpha = 1 + \tan \alpha \]
\[ (5k - 1)\tan \alpha = 5k - 1 \implies \tan \alpha = 1 \quad \text{(if \( 5k - 1 \neq 0 \))} \]

From the second equation:
\[ 3k(1 - \tan \beta) = 1 + \tan \beta \implies 3k - 3k \tan \beta = 1 + \tan \beta \]
\[ (3k - 1)\tan \beta = 3k - 1 \implies \tan \beta = 1 \quad \text{(if \( 3k - 1 \neq 0 \))} \]

From the third equation:
\[ 2k(1 - \tan \gamma) = 1 + \tan \gamma \implies 2k - 2k \tan \gamma = 1 + \tan \gamma \]
\[ (2k - 1)\tan \gamma = 2k - 1 \implies \tan \gamma = 1 \quad \text{(if \( 2k - 1 \neq 0 \))} \]

Thus, we find that:
\[ \tan \alpha = \tan \beta = \tan \gamma = 1 \implies \alpha = \beta = \gamma = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

Now we substitute \( \alpha = \beta = \gamma = \frac{\pi}{4} \) into the expression we need to evaluate:

\[ 12\sin^2(\alpha - \beta) + 15\sin^2(\beta - \gamma) - 7\sin^2(\gamma - \alpha) \]

Since \( \alpha = \beta = \gamma \):
\[ \sin^2(\alpha - \beta) = \sin^2(0) = 0 \]
\[ \sin^2(\beta - \gamma) = \sin^2(0) = 0 \]
\[ \sin^2(\gamma - \alpha) = \sin^2(0) = 0 \]

Thus, the expression simplifies to:
\[ 12 \cdot 0 + 15 \cdot 0 - 7 \cdot 0 = 0 \]

Therefore, the final answer is:

\[ \boxed{0} \]