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Grade: 11
        The option of the question given above is option a. The question is from complex number.
2 years ago

Answers : (1)

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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