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 \( \cos x - \sin \alpha \cdot \cot \beta \cdot \sin x = \cos \alpha \) and find the value of \( \tan\left(\frac{x}{2}\right) \), we can follow these steps:

### Step 1: Rewrite \( \cos x \) and \( \sin x \) in terms of \( \tan\left(\frac{x}{2}\right) \)

Using the half-angle identities:
\[ \cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \]
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