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STD 11 - 4.1 complex nubers — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 4.1 complex nubers1 Mark
MCQ
If $a = \cos \,\theta + i\,\sin \,\theta ,$ then $\frac{{1 + a}}{{1 - a}} = $
  • A
    $\cot \theta $
  • B
    $\cot \frac{\theta }{2}$
  • ✓
    $i\,\cot \frac{\theta }{2}$
  • D
    $i\,\tan \frac{\theta }{2}$
✓

Answer

Correct option: C.
$i\,\cot \frac{\theta }{2}$
c
(c) $a = \cos \theta + i\sin \theta .$
$\frac{{1 + a}}{{1 - a}} = \frac{{(1 + \cos \theta ) + i\sin \theta }}{{(1 - \cos \theta ) - i\sin \theta }}.\,$
Rationalization of denominator, we get $\frac{{1 + a}}{{1 - a}} = \frac{{(1 + \cos \theta ) + i\sin \theta }}{{(1 - \cos \theta ) - i\,\sin \theta }} \times \frac{{(1 - \cos \theta ) + i\sin \theta }}{{(1 - \cos \theta ) + i\sin \theta }}$
$ = \frac{{(1 + \cos \theta )\,(1 - \cos \theta ) + (1 + \cos \theta )\,i\sin \theta + (1 - \cos \theta )i\sin \theta + {i^2}{{\sin }^2}\theta }}{{{{(1 - \cos \theta )}^2} - {{(i\sin \theta )}^2}}}$
$ = \frac{{1 - ({{\cos }^2}\theta + {{\sin }^2}\theta ) + 2i\sin \theta }}{{1 + ({{\cos }^2}\theta + {{\sin }^2}\theta ) - 2\,\cos \theta }}$$ = \frac{{2i\sin \theta }}{{2(1 - \cos \theta )}}$
$ = \frac{{i.2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2{{\sin }^2}\frac{\theta }{2}}}$$ = i\frac{{\cos \frac{\theta }{2}}}{{\sin \frac{\theta }{2}}} = i\cot \frac{\theta }{2}$.

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