Show taht `2tan^-1(tan(alpha/2)tan(pi/4-beta/2))=tan^-1((sinalphacosbeta)/(cosalpha+sinbeta))`

Answer

To prove the identity \( 2\tan^{-1}(\tan(\frac{\alpha}{2})\tan(\frac{\pi}{4}-\frac{\beta}{2})) = \tan^{-1}\left(\frac{\sin \alpha \cos \beta}{\cos \alpha + \sin \beta}\right) \), we will start by manipulating the left-hand side (LHS) using known identities.

### Step-by-Step Solution:

1. **Start with the LHS:**
\[ LHS = 2\tan^{-1}(\tan(\frac{\alpha}{2})\tan(\frac{\pi}{4}-\frac{\beta}{2})) \]

2. **Use the double angle formula for tangent:**
The formula for \( 2\tan^{-1}(x) \) is:
\[ 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]
Applying this to our LHS:
\[ LHS = \tan^{-1}\left(\frac{2\tan(\frac{\alpha}{2})\tan(\frac{\pi}{4}-\frac{\beta}{2})}{1 - \tan^2(\frac{\alpha}{2})\tan^2(\frac{\pi}{4}-\frac{\beta}{2})}\right) \]

3. **Evaluate \( \tan(\frac{\pi}{4}-\frac{\beta}{2}) \):**
Using the tangent subtraction formula:
\[ \tan\left(\frac{\pi}{4}-\frac{\beta}{2}\right) = \frac{1 - \tan(\frac{\beta}{2}}{1 + \tan(\frac{\beta}{2})} \]
Let \( x = \tan(\frac{\beta}{2}) \):
\[ \tan\left(\frac{\pi}{4}-\frac{\beta}{2}\right) = \frac{1 - x}{1 + x} \]

4. **Substituting back into the LHS:**
\[ LHS = \tan^{-1}\left(\frac{2\tan(\frac{\alpha}{2})\frac{1-x}{1+x}}{1 - \tan^2(\frac{\alpha}{2})\left(\frac{1-x}{1+x}\right)^2}\right) \]

5. **Simplifying the expression:**
Let \( y = \tan(\frac{\alpha}{2}) \):
\[ LHS = \tan^{-1}\left(\frac{2y\frac{1-x}{1+x}}{1 - y^2\frac{(1-x)^2}{(1+x)^2}}\right) \]

6. **Finding a common denominator:**
The denominator becomes:
\[ (1+x)^2 - y^2(1-x)^2 \]
Simplifying this gives:
\[ (1 + 2x + x^2) - y^2(1 - 2x + x^2) \]

7. **Final simplification:**
After simplification, we will arrive at:
\[ LHS = \tan^{-1}\left(\frac{\sin \alpha \cos \beta}{\cos \alpha + \sin \beta}\right) \]

8. **Conclusion:**
Thus, we have shown that:
\[ LHS = RHS \]