RAJASTHAN ­ PET Rajasthan PET Solved Paper-2001

  • question_answer
    \[\frac{(\cos \alpha +i\sin \alpha )(\cos \beta +i\sin \beta )}{(\cos \gamma +i\sin \gamma )(\cos \delta +i\sin \delta )}\]is equal to

    A)  \[\cos (\alpha +\beta -\gamma -\delta )-i\sin (\alpha +\beta -\gamma -\delta )\]

    B)  \[\cos (\alpha +\beta -\gamma +\delta )+i\sin (\alpha +\beta -\gamma +\delta )\]

    C)  \[\sin (\alpha +\beta -\gamma -\delta )+i\cos (\alpha +\beta -\gamma -\delta )\]

    D)  \[\cos (\alpha +\beta -\gamma -\delta )+i\sin (\alpha +\beta -\gamma -\delta )\]

    Correct Answer: D

    Solution :

     \[\frac{(\cos \alpha +i\sin \alpha )(\cos \beta +i\sin \beta )}{(\cos \gamma +i\sin \gamma )(\cos \delta +i\sin \delta )}\] \[=\frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta +i(\cos \alpha \sin \beta +\sin \alpha \cos \beta )}{\cos \gamma \cos \delta -\sin \gamma \sin \delta +i(\cos \gamma \sin \delta +\sin \gamma \cos \delta )}\] \[=\frac{\cos (\alpha +\beta )+i\sin (\alpha +\beta )}{\cos (\gamma +\delta )+i\sin (\gamma +\delta )}\] \[=[\cos (\alpha +\beta )+i\sin (\alpha +\beta )]\] \[{{[\cos (\gamma +\delta )+i\sin (\gamma +\delta )]}^{-1}}\] \[=[\cos (\alpha +\beta )+i\sin (\alpha +\beta )][\cos (\gamma +\delta )-\] \[i\sin (\gamma +\delta )]\] \[=\cos (\alpha +\beta -\gamma -\delta )+i\sin (\alpha +\beta -\gamma -\delta )\]


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