A) 1
B) \[\sqrt{2}\]
C) \[\sqrt{3}\]
D) None of these
Correct Answer: C
Solution :
[c] : Given, \[\cos \alpha =\frac{2\cos \beta -1}{2-\cos \beta }\] \[\Rightarrow \]\[\frac{1-{{\tan }^{2}}\frac{\alpha }{2}}{1+{{\tan }^{2}}\frac{\alpha }{2}}=\frac{2\left( 1-{{\tan }^{2}}\frac{\beta }{2} \right)-\left( 1+{{\tan }^{2}}\frac{\beta }{2} \right)}{2\left( 1+{{\tan }^{2}}\frac{\beta }{2} \right)-\left( 1-{{\tan }^{2}}\frac{\beta }{2} \right)}\] \[\Rightarrow \]\[\frac{1-{{\tan }^{2}}\frac{\alpha }{2}}{1+{{\tan }^{2}}\frac{\alpha }{2}}=\frac{1-3{{\tan }^{2}}\frac{\beta }{2}}{1+3{{\tan }^{2}}\frac{\beta }{2}}\] Applying component and dividendo, we get \[{{\tan }^{2}}\frac{\alpha }{2}=3{{\tan }^{2}}\frac{\beta }{2}\Rightarrow \tan \frac{\alpha }{2}.\cot \frac{\beta }{2}=\sqrt{3}\]
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