2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main - Mock Test - 15

  • question_answer
    If \[\alpha \]and \[\beta \]are solutions of \[{{\sin }^{2}}x+a(sin\,x)+b=0\]as well as that of \[{{\cos }^{2}}x+c(cos\,x)+d=0\], then \[\sin (\alpha +\beta )\] is equal to

    A) \[\frac{2bd}{{{b}^{2}}+{{d}^{2}}}\]

    B) \[\frac{{{a}^{2}}+{{c}^{2}}}{2ac}\]

    C) \[\frac{{{b}^{2}}+{{d}^{2}}}{2bd}\]

    D) \[\frac{2ac}{{{a}^{2}}+{{c}^{2}}}\]

    Correct Answer: D

    Solution :

    [d]: According to the given condition, \[\Rightarrow \]\[2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=-a\] and\[2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=-c\Rightarrow \tan \frac{\alpha +\beta }{2}=\frac{a}{c}\] Now,\[\sin (\alpha +\beta )=\frac{2\tan \left( \frac{\alpha +\beta }{2} \right)}{1+{{\tan }^{2}}\left( \frac{\alpha +\beta }{2} \right)}=\frac{2ac}{{{a}^{2}}+{{c}^{2}}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner