If cos⁡x−cos⁡αcos⁡x−cos⁡β=sin2αcos⁡βsin2⁡βcos⁡α, then
1. A

$\mathrm{cos}x=\frac{2\mathrm{cos}\alpha +\mathrm{cos}\beta }{1+\mathrm{cos}\alpha \mathrm{cos}\beta }$

2. B

$\mathrm{cos}x=\frac{\mathrm{cos}a+\mathrm{cos}\beta }{1-\mathrm{cos}\alpha \mathrm{cos}\beta }$

3. C

$\mathrm{tan}\frac{x}{2}=\mathrm{tan}\frac{\alpha }{2}\mathrm{tan}\frac{\beta }{2}$

4. D

$\mathrm{tan}\frac{x}{2}=-2\mathrm{tan}\frac{\alpha }{2}\mathrm{tan}\frac{\beta }{2}$

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### Solution:

From the given results

$\begin{array}{l}⇒\frac{1-\mathrm{cos}x}{1+\mathrm{cos}x}=\frac{1+\mathrm{cos}\alpha \mathrm{cos}\beta -\mathrm{cos}\alpha -\mathrm{cos}\beta }{1+\mathrm{cos}\alpha \mathrm{cos}\beta +\mathrm{cos}\alpha +\mathrm{cos}\beta }\\ =\frac{\left(1-\mathrm{cos}\alpha \right)\left(1-\mathrm{cos}\beta \right)}{\left(1+\mathrm{cos}\alpha \right)\left(1+\mathrm{cos}\beta \right)}\\ ⇒{\mathrm{tan}}^{2}\frac{x}{2}={\mathrm{tan}}^{2}\frac{\alpha }{2}\cdot {\mathrm{tan}}^{2}\frac{\beta }{2}⇒\mathrm{tan}\frac{x}{2}=±\mathrm{tan}\frac{\alpha }{2}\mathrm{tan}\frac{\beta }{2}\\ ⇒\mathrm{tan}\frac{x}{2}=±\mathrm{tan}\frac{\alpha }{2}\mathrm{tan}\frac{\beta }{2}\end{array}$

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