If `cosx-sinalphacotbetasinx=cosa ,` then the value of `tan(x/2)` is (a)`-tan(alpha/2)cot(beta/2)` (b) `tan(alpha/2)tan(beta/2)` (c)`-cot((alphabeta)/2)tan(beta/2)` (d) `cot(alpha/2)cot(beta/2)`

Explanation

To solve the equation \( \cos x - \sin \alpha \cot \beta \sin x = \cos \alpha \) for the value of \( \tan \left( \frac{x}{2} \right) \), we will follow these steps:

### Step 1: Rewrite the trigonometric functions
We can express \( \cos x \) and \( \sin x \) in terms of \( \tan \left( \frac{x}{2} \right) \):
\[ \cos x = \frac{1 - \tan^2 \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} \]
\[
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