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प्रश्न
If \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\] , prove that \[\tan\frac{x}{2} = \pm \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\]
उत्तर
Given: \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\] ...(1)
\[\Rightarrow \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{cos\alpha + cos\beta}{1 + cos\alpha \times cos\beta} \left[ \because \text{ cos } x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right]\]
\[\text{ By componendo and dividendo, we get } \]
\[\frac{\left( 1 - \tan^2 \frac{x}{2} \right) + \left( 1 + \tan^2 \frac{x}{2} \right)}{\left( 1 - \tan^2 \frac{x}{2} \right) - \left( 1 + \tan^2 \frac{x}{2} \right)} = \frac{\left( 1 + cos\alpha \times cos\beta + cos\alpha + cos\beta \right)}{- \left( 1 + cos\alpha cos\beta - cos\alpha - cos\beta \right)}\]
\[ \Rightarrow \frac{2}{2 \tan^2 \frac{x}{2}} = \frac{\left( 1 + cos\alpha \right)\left( 1 + cos\beta \right)}{\left( 1 - cos\alpha \right)\left( 1 - cos\beta \right)}\]
\[\Rightarrow \tan^2 \frac{x}{2} = \frac{\left( 1 - cos\alpha \right)\left( 1 - cos\beta \right)}{\left( 1 + cos\alpha \right)\left( 1 + cos\beta \right)}\]
\[ \Rightarrow \tan^2 \frac{x}{2} = \frac{2 \sin^2 \frac{\alpha}{2} \times 2 \sin^2 \frac{\beta}{2}}{2 \cos^2 \frac{\alpha}{2} \times 2 \cos^2 \frac{\beta}{2}}\]
\[ \Rightarrow \tan^2 \frac{x}{2} = \tan^2 \frac{\alpha}{2} \times \tan^2 \frac{\beta}{2}\]
\[ \Rightarrow \tan\frac{x}{2} = \pm \tan\frac{\alpha}{2} \times \tan\frac{\beta}{2}\]
\[\text{ Hence proved } .\]
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