If `tan alpha = (1)/(sqrt (x (x ^(2) + x + 1))). tan beta = (sqrtx)/(sqrt (x ^(2) + x +1))and tan y = (x ^(-3) + x ^(-2)+ x ^(-1)) ^(1//2),` then for `x gt 0`

Options

A. `alpha + beta = 2 gamma `
B. `alpha +beta = 3 gamma `
C. `alpha + beta = gamma`
D. `2 (alpha + beta) = gamma `

Answer

To solve the problem, we need to find the relationship between the angles α, β, and γ given their tangent values. Let's break it down step by step.

### Step 1: Write down the given tangent values
We have:
- \( \tan \alpha = \frac{1}{\sqrt{x(x^2 + x + 1)}} \)
- \( \tan \beta = \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}} \)
- \( \tan \gamma = \sqrt{x^{-3} + x^{-2} + x^{-1}} \)

### Step 2: Use the tangent addition formula
To find \( \tan(\alpha + \beta) \), we use the formula:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

### Step 3: Calculate \( \tan \alpha + \tan \beta \)
Substituting the values of \( \tan \alpha \) and \( \tan \beta \):
\[ \tan \alpha + \tan \beta = \frac{1}{\sqrt{x(x^2 + x + 1)}} + \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}} \]
Finding a common denominator:
\[ = \frac{1 + x}{\sqrt{x(x^2 + x + 1)}} \]

### Step 4: Calculate \( \tan \alpha \tan \beta \)
Now calculate \( \tan \alpha \tan \beta \):
\[ \tan \alpha \tan \beta = \left(\frac{1}{\sqrt{x(x^2 + x + 1)}}\right) \left(\frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}\right) = \frac{1}{x^2 + x + 1} \]

### Step 5: Substitute into the tangent addition formula
Now substitute back into the tangent addition formula:
\[ \tan(\alpha + \beta) = \frac{\frac{1 + x}{\sqrt{x(x^2 + x + 1)}}}{1 - \frac{1}{x^2 + x + 1}} \]

### Step 6: Simplify the denominator
The denominator simplifies as follows:
\[ 1 - \frac{1}{x^2 + x + 1} = \frac{(x^2 + x + 1) - 1}{x^2 + x + 1} = \frac{x^2 + x}{x^2 + x + 1} \]

### Step 7: Combine the expressions
Now, we can write:
\[ \tan(\alpha + \beta) = \frac{(1 + x)(x^2 + x + 1)}{\sqrt{x(x^2 + x + 1)}(x^2 + x)} \]

### Step 8: Simplify further
This expression can be simplified further, but we need to relate it to \( \tan \gamma \):
\[ \tan \gamma = \sqrt{x^{-3} + x^{-2} + x^{-1}} = \sqrt{\frac{1}{x^3} + \frac{1}{x^2} + \frac{1}{x}} = \sqrt{\frac{1 + x + x^2}{x^3}} = \frac{\sqrt{1 + x + x^2}}{x^{3/2}} \]

### Step 9: Equate \( \tan(\alpha + \beta) \) and \( \tan \gamma \)
We need to show that:
\[ \tan(\alpha + \beta) = \tan \gamma \]
This leads us to conclude that:
\[ \alpha + \beta = \gamma \]

### Final Conclusion
Thus, the relation between the angles is:
\[ \alpha + \beta = \gamma \]