The value `2tan^(-1)[sqrt((a-b)/(a+b)tantheta/2)]` is equal to `cos^(-1)((acostheta+b)/(a+bcostheta))` (b) `cos^(-1)((a+bcostheta)/(acostheta+b))` `cos^(-1)((acostheta)/(a+bcostheta))` (d) `cos^(-1)((bcostheta)/(acostheta+b))`

Options

A. `cos^(-1) ((a cos theta + b)/(a + b cos theta))`
B. `cos^(-1) ((a + b cos theta)/(a cos theta + b))`
C. `cos^(-1) ((a cos theta)/(a + b cos theta))`
D. `cos^(-1) ((b cos theta)/(a cos theta + b))`

Explanation

To solve the problem, we need to evaluate the expression \(2\tan^{-1}\left(\sqrt{\frac{(a-b)}{(a+b)}\tan\frac{\theta}{2}}\right)\) and show that it is equal to one of the given options.

### Step-by-step Solution:

1. **Start with the given expression:**
\[ 2\tan^{-1}\left(\sqrt{\frac{(a-b)}{(a+b)}\tan\frac{\theta}{2}}\right) \]

2. **Use the double angle formula for tangent:**
Recall that:
\[ \tan(2x) = \frac{2\tan x}{1 - \tan^2 x} \]
Let \(x = \tan^{-1}\left(\sqrt{\frac{(a-b)}{(a+b)}\tan\frac{\theta}{2}}\right)\). Then:
\[ \tan(2x) = \frac{2\sqrt{\frac{(a-b)}{(a+b)}\tan\frac{\theta}{2}}}{1 - \left(\frac{(a-b)}{(a+b)}\tan\frac{\theta}{2}\right)} \]

3. **Simplify the expression:**
Substitute \(u = \tan\frac{\theta}{2}\):
\[ \tan(2x) = \frac{2\sqrt{\frac{(a-b)}{(a+b)}u}}{1 - \frac{(a-b)}{(a+b)}u} \]

4. **Convert to cosine:**
We can express this in terms of cosine using the identity:
\[ \tan(2x) = \frac{\sin(2x)}{\cos(2x)} \]
and relate it to cosine:
\[ 2x = \cos^{-1}\left(\frac{1 - \tan^2 x}{1 + \tan^2 x}\right) \]

5. **Using the tangent half-angle identity:**
We know:
\[ \tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta} \]
Substitute this into our expression.

6. **Final expression:**
After simplification, we arrive at:
\[ \cos^{-1}\left(\frac{a \cos \theta + b}{a + b \cos \theta}\right) \]

7. **Choose the correct option:**
Comparing with the options provided, we find that:
\[ 2\tan^{-1}\left(\sqrt{\frac{(a-b)}{(a+b)}\tan\frac{\theta}{2}}\right) = \cos^{-1}\left(\frac{a \cos \theta + b}{a + b \cos \theta}\right) \]
This corresponds to option (a).

### Conclusion:
The value \(2\tan^{-1}\left(\sqrt{\frac{(a-b)}{(a+b)}\tan\frac{\theta}{2}}\right)\) is equal to:
\[ \cos^{-1}\left(\frac{a \cos \theta + b}{a + b \cos \theta}\right) \]