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The option of the question given above is option a. The question is from complex number.

The option of the question given above is option a. The question is from complex number.

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Grade:11

1 Answers

Arun
25750 Points
6 years ago
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

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