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

  • question_answer         If tan \[\theta =\frac{b}{a}\]then the value of \[a.cos2\theta +b.sin2\theta \] is

    A) a                     

    B) b              

    C) \[a-b\]                           

    D) \[a+b\]

    Correct Answer: A

    Solution :

    [a] \[a.cos2\theta +b.sin2\theta \] \[=a.\frac{1-tan\theta }{1+ta{{n}^{2}}\theta }+b\frac{2tan2\theta }{1+ta{{n}^{2}}\theta }=\frac{a(1-ta{{n}^{2}}\theta )+2btan\theta }{1+ta{{n}^{2}}\theta }\] \[=\frac{a\left( 1-\frac{{{b}^{2}}}{{{a}^{2}}} \right)+2b.\frac{b}{a}}{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\frac{a({{a}^{2}}-{{b}^{2}})+2a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[=\frac{{{a}^{3}}-a{{b}^{2}}+2a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[=\frac{{{a}^{3}}+a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=a\] Hence, option [a] is correct.

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