If `atanalpha+sqrt(a^2-1)tanbeta+sqrt(a^2+1)tangamma=2a` where a is a constant and `alpha,beta and gamma` are variable angles. Then the least value of `tan^2alpha+tan^2beta+tan^2gamma` is equal to

Answer

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

\[ \tan \alpha + \sqrt{a^2 - 1} \tan \beta + \sqrt{a^2 + 1} \tan \gamma = 2a \]

where \( a \) is a constant and \( \alpha, \beta, \gamma \) are variable angles. We need to find the least value of \( \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma \).

### Step 1: Rewrite the equation in vector form

We can express the equation in a vector form. Let:

\[ \mathbf{A} = a \hat{i} + \sqrt{a^2 - 1} \hat{j} + \sqrt{a^2 + 1} \hat{k} \]

and

\[ \mathbf{B} = \tan \alpha \hat{i} + \tan \beta \hat{j} + \tan \gamma \hat{k} \]

The equation then becomes:

\[ \mathbf{A} \cdot \mathbf{B} = 2a \]

### Step 2: Calculate the magnitudes of vectors

The magnitude of vector \( \mathbf{A} \) is:

\[ |\mathbf{A}| = \sqrt{a^2 + (a^2 - 1) + (a^2 + 1)} = \sqrt{3a^2} = \sqrt{3}a \]

The magnitude of vector \( \mathbf{B} \) is:

\[ |\mathbf{B}| = \sqrt{\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma} \]

### Step 3: Use the dot product formula

Using the dot product formula:

\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]

we can write:

\[ \sqrt{3}a \sqrt{\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma} \cos \theta = 2a \]

### Step 4: Simplify the equation

Dividing both sides by \( a \) (assuming \( a \neq 0 \)) gives:

\[ \sqrt{3} \sqrt{\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma} \cos \theta = 2 \]

### Step 5: Isolate the square root

Now, isolating \( \sqrt{\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma} \):

\[ \sqrt{\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma} = \frac{2}{\sqrt{3} \cos \theta} \]

### Step 6: Square both sides

Squaring both sides gives:

\[ \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma = \frac{4}{3 \cos^2 \theta} \]

### Step 7: Find the minimum value

The minimum value of \( \sec^2 \theta \) is 1 (when \( \cos \theta = 1 \)), thus:

\[ \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma \geq \frac{4}{3} \]

### Conclusion

Therefore, the least value of \( \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma \) is:

\[ \boxed{\frac{4}{3}} \]