SSC Sample Paper Mock Test-11 SSC CGL Tear-II Paper-1

  • question_answer
    If 9 lies in the third quadrant and \[3\tan \theta -4=0,\]then \[5\sin 2\theta +3\sin \theta +4\cos \theta \] is equal to

    A)  \[0\]    

    B)  \[\frac{-\,\,24}{5}\]

    C)  \[\frac{24}{5}\]

    D)  \[\frac{48}{5}\]

    Correct Answer: A

    Solution :

    \[3\tan \theta =4\]\[\Rightarrow \]\[\sin \theta =\frac{-\,\,4}{5}\]and \[\cos \theta =\frac{-\,\,3}{5}\]
    \[\therefore \] \[5\sin 2\theta +3\sin \theta +4\cos \theta \]  
    \[=10\sin \theta \cos \theta +3\sin \theta +4\cos \theta \]
    \[=10\left( \frac{-\,\,4}{5} \right)\left( \frac{-\,\,3}{5} \right)+3\left( \frac{-\,\,4}{5} \right)+4\left( \frac{-\,\,3}{5} \right)\]
    \[=10\left( \frac{12}{25} \right)-\frac{12}{5}-\frac{12}{5}=\frac{24}{5}-\frac{24}{5}=0\]


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