JEE Main & Advanced Sample Paper JEE Main - Mock Test - 21

  • question_answer
    If\[\cos \alpha =\frac{2\cos \beta -1}{2-\cos \beta },(0<\alpha <\pi ,0<\beta <\pi )\], then \[\tan \frac{\alpha }{2}\cot \frac{\beta }{2}\] is equal to

    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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