11th Class Mathematics Sample Paper Mathematics Olympiad - Sample Paper-4

  • question_answer If \[\mathbf{tanA}=\left( \frac{1-\mathbf{cosB}}{\sin B} \right)\]then tan2A is equal to:

    A) \[tanB\]                        

    B) \[tan\frac{B}{2}\]         

    C) \[\frac{\tan B}{2}\]                               

    D) None of these

    Correct Answer: A

    Solution :

    [a]\[\tan A=\frac{1-cosB}{\sin B}\] \[tan2A=\frac{2tanA}{1-{{\tan }^{2}}A}=\frac{2\frac{(1-cosB)}{sinB}}{\frac{1-{{(1-cosB)}^{2}}}{{{\sin }^{2}}B}}=\frac{2\left( 1-\cos B \right).sinB}{{{\sin }^{2}}B-{{\left( 1-\cos B \right)}^{2}}}\] \[=\frac{2sinB\left( 1-\cos B \right)}{{{\left( 1-\cos B \right)}^{2}}-{{\left( 1-\cos B \right)}^{2}}}=\frac{2sinB\left( 1-\cos B \right)}{\left( 1-\cos B \right)-[1+\cos B-1+\cos B]}\] \[=\frac{2\sin B}{2\cos B}=\tan B.\]

adversite


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