JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Integral power of iota, Algebraic operations and Equality of complex numbers

  • question_answer The real part of  \[{{(1-\cos \theta +2i\sin \theta )}^{-1}}\]is [IIT 1978, 86]

    A) \[\frac{1}{3+5\cos \theta }\]

    B) \[\frac{1}{5-3\cos \theta }\]

    C) \[\frac{1}{3-5\cos \theta }\]

    D) \[\frac{1}{5+3\cos \theta }\]

    Correct Answer: D

    Solution :

    \[{{\{(1-\cos \theta )+i.2\sin \theta \}}^{-1}}={{\left\{ 2{{\sin }^{2}}\frac{\theta }{2}+i.4\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right\}}^{-1}}\] = \[{{\left( 2\sin \frac{\theta }{2} \right)}^{-1}}{{\left\{ \sin \frac{\theta }{2}+i.2\cos \frac{\theta }{2} \right\}}^{-1}}\] \[={{\left( 2\sin \frac{\theta }{2} \right)}^{-1}}\frac{1}{\sin \frac{\theta }{2}+i.2\cos \frac{\theta }{2}}\times \frac{\sin \frac{\theta }{2}-i.2\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}-i.2\cos \frac{\theta }{2}}\] \[=\frac{\sin \frac{\theta }{2}-i.2\cos \frac{\theta }{2}}{2\sin \frac{\theta }{2}\left( {{\sin }^{2}}\frac{\theta }{2}+4{{\cos }^{2}}\frac{\theta }{2} \right)}\]. it?s real part     \[=\frac{\sin \frac{\theta }{2}}{2\sin \frac{\theta }{2}\left( 1+3{{\cos }^{2}}\frac{\theta }{2} \right)}=\frac{1}{2\left( 1+3{{\cos }^{2}}\frac{\theta }{2} \right)}\]\[=\frac{1}{5+3\cos \theta }\]

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