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`        The option of the question given above is option a. The question is from complex number.`
2 years ago

Arun
23742 Points
```							Dear Subodh For the part in bracket Multiply and divide by [(1+sin@) + i cos@]  [(1 + sin@) + i cos @]² / [(1+sin@)² - (i cos@)²]= [1 + sin²@ + 2sin@ - cos²@ + 2i(cos@ + sin@* cos@)]/ [1 + sin²@ + 2sin@ + cos²@] = [2sin@(1+ sin@) + 2i cos@(1 + sin@)]/ 2(1 + sin@)= Sin@ + i cos@= Cos (pi/2 - @) + i sin(pi/2 -@)Now question becomes[Cos(pi/2-@) + i sin(pi/2- @)]n= cos (n*pi/2 - n@) + i sin ( n*pi/2 -n@) Hence option A is correct
```
2 years ago
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