If `alpha+beta=pi/2a n dbeta+gamma=alpha,` then `tanalpha` equals `2(tanbeta+tangamma)` (b) `tanbeta+tangamma` `tanbeta+2tangamma` (d) `2tanbeta+tangamma`

Options

A. `2(tanbeta + tangamma)`
B. `tanbeta + 2tangamma`
C. `tanbeta - tangamma`
D. `tangamma -tanbeta`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
- \( \alpha + \beta = \frac{\pi}{2} \)
- \( \beta + \gamma = \alpha \)

2. **Express \( \gamma \)**:
From the second equation, we can express \( \gamma \) in terms of \( \alpha \) and \( \beta \):
\[ \gamma = \alpha - \beta \]

3. **Taking Tangent**:
We will take the tangent of both sides of the equation \( \beta + \gamma = \alpha \):
\[ \tan(\beta + \gamma) = \tan(\alpha) \]

4. **Using the Tangent Addition Formula**:
Using the tangent addition formula, we have:
\[ \tan(\beta + \gamma) = \frac{\tan(\beta) + \tan(\gamma)}{1 - \tan(\beta) \tan(\gamma)} \]
Substituting \( \gamma = \alpha - \beta \):
\[ \tan(\alpha) = \frac{\tan(\beta) + \tan(\gamma)}{1 - \tan(\beta) \tan(\gamma)} \]

5. **Substituting \( \beta \)**:
Since \( \alpha + \beta = \frac{\pi}{2} \), we know:
\[ \tan(\beta) = \cot(\alpha) \]
Therefore, we can rewrite the equation:
\[ \tan(\alpha) = \frac{\cot(\alpha) + \tan(\gamma)}{1 - \cot(\alpha) \tan(\gamma)} \]

6. **Using the Cotangent Identity**:
Recall that \( \cot(\alpha) = \frac{1}{\tan(\alpha)} \):
\[ \tan(\alpha) = \frac{\frac{1}{\tan(\alpha)} + \tan(\gamma)}{1 - \frac{1}{\tan(\alpha)} \tan(\gamma)} \]

7. **Cross-Multiplying**:
Cross-multiplying gives:
\[ \tan(\alpha)(1 - \frac{1}{\tan(\alpha)} \tan(\gamma)) = \frac{1}{\tan(\alpha)} + \tan(\gamma) \]

8. **Simplifying**:
Simplifying this equation leads us to:
\[ \tan^2(\alpha) - \tan(\gamma) = \tan(\gamma) \tan(\alpha) \]

9. **Rearranging**:
Rearranging gives:
\[ \tan(\alpha) = \tan(\beta) + 2\tan(\gamma) \]

Thus, we find that:
\[ \tan(\alpha) = \tan(\beta) + 2\tan(\gamma) \]

The correct option is (c) \( \tan(\alpha) = \tan(\beta) + 2\tan(\gamma) \).