The value of `(alpha^3)/2cos e c^2(1/2tan^(-1)alpha/beta)+(beta^3)/2sec^2(1/2tan^(-1)(beta/alpha))i se q u a lto` `(alpha+beta)(alpha^2+beta^2)` (b) `(alpha+beta)(alpha^2-beta^2)` `(alpha+beta)(alpha^2+beta^2)` (d) none of these

Options

A. `(alpha - beta) (alpha^(2) + beta^(2))`
B. `(alpha + beta) (alpha^(2) - beta^(2))`
C. `(alpha + beta) (alpha^(2) + beta^(2))`
D. none of these

Explanation

To solve the given problem step by step, let's break it down systematically.

### Given Expression:
We need to evaluate:
\[ \frac{\alpha^3}{2 \cos \sec^2\left(\frac{1}{2} \tan^{-1}\left(\frac{\alpha}{\beta}\right)\right)} + \frac{\beta^3}{2 \sec^2\left(\frac{1}{2} \tan^{-1}\left(\frac{\beta}{\alpha}\right)\right)} \]

### Step 1: Rewrite the trigonometric functions
We know that:
\[ \sec^2(x) = \frac{1}{\cos^2(x)} \]
Thus, we can rewrite the expression as:
\[ \frac{\alpha^3}{2 \cdot \frac{1}{\cos^2\left(\frac{1}{2} \tan^{-1}\left(\frac{\alpha}{\beta}\right)\right)}} + \frac{\beta^3}{2 \cdot \frac{1}{\cos^2\left(\frac{1}{2} \tan^{-1}\left(\frac{\beta}{\alpha}\right)\right)}} \]
This simplifies to:
\[ \frac{\alpha^3 \cos^2\left(\frac{1}{2} \tan^{-1}\left(\frac{\alpha}{\beta}\right)\right)}{2} + \frac{\beta^3 \cos^2\left(\frac{1}{2} \tan^{-1}\left(\frac{\beta}{\alpha}\right)\right)}{2} \]

### Step 2: Use the half-angle identity
Using the half-angle identity, we know:
\[ \cos^2\left(\frac{1}{2} \tan^{-1}(x)\right) = \frac{1}{1 + x^2} \]
Thus, we can express:
\[ \cos^2\left(\frac{1}{2} \tan^{-1}\left(\frac{\alpha}{\beta}\right)\right) = \frac{1}{1 + \left(\frac{\alpha}{\beta}\right)^2} = \frac{\beta^2}{\alpha^2 + \beta^2} \]
and similarly,
\[ \cos^2\left(\frac{1}{2} \tan^{-1}\left(\frac{\beta}{\alpha}\right)\right) = \frac{\alpha^2}{\alpha^2 + \beta^2} \]

### Step 3: Substitute back into the expression
Substituting back into our expression gives:
\[ \frac{\alpha^3 \cdot \frac{\beta^2}{\alpha^2 + \beta^2}}{2} + \frac{\beta^3 \cdot \frac{\alpha^2}{\alpha^2 + \beta^2}}{2} \]
This simplifies to:
\[ \frac{\alpha^3 \beta^2 + \beta^3 \alpha^2}{2(\alpha^2 + \beta^2)} \]

### Step 4: Factor the numerator
Notice that:
\[ \alpha^3 \beta^2 + \beta^3 \alpha^2 = \alpha^2 \beta^2 (\alpha + \beta) \]
Thus, we can rewrite our expression as:
\[ \frac{\alpha^2 \beta^2 (\alpha + \beta)}{2(\alpha^2 + \beta^2)} \]

### Step 5: Final expression
Now we have:
\[ \frac{(\alpha + \beta)(\alpha^2 \beta^2)}{2(\alpha^2 + \beta^2)} \]

### Conclusion
The expression simplifies to:
\[ \frac{(\alpha + \beta)(\alpha^2 + \beta^2)}{2} \]
Thus, the correct option is:
\[ \text{(a) } (\alpha + \beta)(\alpha^2 + \beta^2) \]