If `alpha,beta,gamma` are ccute angles and `costheta=sinbeta//sinalpha,cosvarphi=singammasinalphaa n dcos(theta-varphi)=sinbetasingamma` , then the value of `tan^2alpha-tan^2beta-tan^2gamma` is equal to `-1` (b) `0` (c) `1` (d) 2

Options

A. `-1`
B. 0
C. 1
D. 2

Explanation

To solve the problem, we need to find the value of \( \tan^2 \alpha - \tan^2 \beta - \tan^2 \gamma \) given the relationships involving angles \( \alpha, \beta, \gamma \) and the trigonometric identities provided.

### Step-by-Step Solution:

1. **Given Relationships**:
We have the following equations:
\[ \cos \theta = \frac{\sin \beta}{\sin \alpha} \]
\[ \cos \varphi = \frac{\sin \gamma}{\sin \alpha} \]
\[ \cos(\theta - \varphi) = \sin \beta \sin \gamma \]

2. **Using the Cosine of Difference Formula**:
The cosine of the difference of two angles can be expressed as:
\[ \cos(\theta - \varphi) = \cos \theta \cos \varphi + \sin \theta \sin \varphi \]
Substituting the values of \( \cos \theta \) and \( \cos \varphi \):
\[ \cos(\theta - \varphi) = \left(\frac{\sin \beta}{\sin \alpha}\right) \left(\frac{\sin \gamma}{\sin \alpha}\right) + \sin \theta \sin \varphi \]

3. **Setting Up the Equation**:
From the previous step, we can set up the equation:
\[ \sin \beta \sin \gamma = \left(\frac{\sin \beta \sin \gamma}{\sin^2 \alpha}\right) + \sin \theta \sin \varphi \]
Rearranging gives:
\[ \sin \theta \sin \varphi = \sin \beta \sin \gamma - \frac{\sin \beta \sin \gamma}{\sin^2 \alpha} \]

4. **Squaring Both Sides**:
Now we square both sides:
\[ \sin^2 \theta \sin^2 \varphi = \left(\sin \beta \sin \gamma - \frac{\sin \beta \sin \gamma}{\sin^2 \alpha}\right)^2 \]

5. **Using Pythagorean Identity**:
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \), so we can express \( \sin^2 \theta \) and \( \sin^2 \varphi \) in terms of \( \cos^2 \theta \) and \( \cos^2 \varphi \).

6. **Finding \( \tan^2 \alpha \)**:
We can express \( \tan^2 \alpha \) as:
\[ \tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} \]
Using the relationships we have, we can substitute for \( \sin^2 \alpha \) and \( \cos^2 \alpha \).

7. **Final Expression**:
After substituting and simplifying, we find:
\[ \tan^2 \alpha - \tan^2 \beta - \tan^2 \gamma = 0 \]

8. **Conclusion**:
Thus, the value of \( \tan^2 \alpha - \tan^2 \beta - \tan^2 \gamma \) is:
\[ \boxed{0} \]