2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Trigonometric Identities

JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Self Evaluation Test - Trigonometric Function

  • question_answer
    If A and B are positive acute angles satisfying \[3{{\cos }^{2}}A+2{{\cos }^{2}}B=4\] and \[\frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A}\] Then the value of \[A+2B\]is equal to:

    A) \[\frac{\pi }{6}\]            

    B) \[\frac{\pi }{2}\]   

    C) \[\frac{\pi }{3}\]            

    D) \[\frac{\pi }{4}\]

    Correct Answer: B

    Solution :

    Given, \[3{{\cos }^{2}}A+2{{\cos }^{2}}B=4\] \[\Rightarrow \,\,2{{\cos }^{2}}B-1=4-3{{\cos }^{2}}A-1\] \[\Rightarrow \,\,\cos 2B=3(1-{{\cos }^{2}}A)=3{{\sin }^{2}}A\]           ?.(1) and \[2\cos B\,\sin B=3\sin A\,\cos A\] \[\sin 2B=3\sin A\cos A\]                         ?..(2) Now, \[\cos (A+2B)=\cos A\cos 2B-\sin A\sin 2B\] \[=\cos A(3{{\sin }^{2}}A)-\sin A(3sin\,A\,\,cos\,A)=0\]                         [using eqs.(1) and (2)] \[\Rightarrow \,\,\,A+2B=\frac{\pi }{2}\]


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