`2tan^(-1){tan.(alpha)/2tan(pi/4-beta/2)}" is equal to " alpha,beta,in (0,pi/2)`

Options

A. `tan^(-1)((cosalpha=sinbeta)/(sinalphacosbeta))`
B. `tan^(-1)((sinalphacosbeta)/(cosalpha+sinbeta))`
C. Both (1) & (2)
D. `tan^(-1)((tanalphatanbeta)/(tanalpha+tanbeta))`

Answer

To solve the equation \( 2\tan^{-1}\left(\frac{\tan(\alpha)}{2\tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right)}\right) = \alpha \), where \( \alpha, \beta \in (0, \frac{\pi}{2}) \), we can follow these steps:

### Step 1: Use the Double Angle Formula for Tangent
We know that:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Let \( \theta = \tan^{-1}\left(\frac{\tan(\alpha)}{2\tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right)}\right) \). Then we can rewrite the left-hand side:
\[ 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \]
where \( x = \frac{\tan(\alpha)}{2\tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right)} \).

### Step 2: Substitute for \( x \)
Substituting \( x \) into the formula gives:
\[ \tan(2\tan^{-1}(x)) = \frac{2\left(\frac{\tan(\alpha)}{2\tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right)}\right)}{1 - \left(\frac{\tan(\alpha)}{2\tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right)}\right)^2} \]

### Step 3: Simplify the Expression
This simplifies to:
\[ \tan(2\tan^{-1}(x)) = \frac{\tan(\alpha)}{\tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right)} \cdot \frac{1 - \frac{\tan^2(\alpha)}{4\tan^2\left(\frac{\pi}{4} - \frac{\beta}{2}\right)}}{1} \]

### Step 4: Use the Identity for \( \tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right) \)
Using the identity:
\[ \tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan(x)}{1 + \tan(x)} \]
we can express \( \tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right) \) in terms of \( \tan\left(\frac{\beta}{2}\right) \).

### Step 5: Set the Equation Equal to \( \tan(\alpha) \)
Now we set:
\[ \tan(2\tan^{-1}(x)) = \tan(\alpha) \]
This leads us to the equation:
\[ \frac{\tan(\alpha)}{\tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right)} = \tan(\alpha) \]

### Step 6: Solve for \( \beta \)
From this equation, we can isolate \( \beta \) and find its relationship with \( \alpha \).

### Final Result
After simplifying and solving, we find:
\[ \tan(\beta) = \tan(\alpha) \cdot \tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right) \]
This leads us to the conclusion that:
\[ \beta = \tan^{-1}\left(\frac{\tan(\alpha)}{1 + \tan(\alpha)}\right) \]

Thus, the final answer is:
\[ \alpha, \beta \in (0, \frac{\pi}{2}) \]