If a, b, c and k are real constants and `alpha, beta, gamma` are variables subject to the condition that `a tanalpha + btanbeta+ c tangamma=k` , then prove using vectors that `tan^2 alpha + tan^2 beta+ tan^2gamma>= k^2 /(a^2 + b^2 + c^2)`

Answer

To prove that \( \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma \geq \frac{k^2}{a^2 + b^2 + c^2} \) given the condition \( a \tan \alpha + b \tan \beta + c \tan \gamma = k \), we will use the concept of vectors.

### Step-by-step Solution:

1. **Define Vectors**:
Let \( \mathbf{u} = (a, b, c) \) and \( \mathbf{v} = (\tan \alpha, \tan \beta, \tan \gamma) \).

2. **Dot Product Representation**:
The dot product of these vectors can be expressed as:
\[ \mathbf{u} \cdot \mathbf{v} = a \tan \alpha + b \tan \beta + c \tan \gamma = k \]

3. **Magnitude of Vectors**:
The magnitudes of the vectors are:
\[ |\mathbf{u}| = \sqrt{a^2 + b^2 + c^2} \]
\[ |\mathbf{v}| = \sqrt{\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma} \]

4. **Cosine of the Angle**:
The cosine of the angle \( \theta \) between the vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by:
\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \]

5. **Substituting the Dot Product**:
Substituting the dot product into the cosine formula gives:
\[ \cos \theta = \frac{k}{\sqrt{a^2 + b^2 + c^2} \cdot \sqrt{\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma}} \]

6. **Squaring Both Sides**:
Squaring both sides results in:
\[ \cos^2 \theta = \frac{k^2}{(a^2 + b^2 + c^2)(\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma)} \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ (a^2 + b^2 + c^2) \cos^2 \theta = \frac{k^2}{\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma} \]

8. **Considering the Range of Cosine**:
Since \( \cos^2 \theta \) is always between 0 and 1, we have:
\[ (a^2 + b^2 + c^2) \cos^2 \theta \leq a^2 + b^2 + c^2 \]

9. **Final Inequality**:
Thus, we can conclude:
\[ \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma \geq \frac{k^2}{a^2 + b^2 + c^2} \]