Plz help ... No. 17
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1
if u look at it carefully u will get a symmetry
A=cosθ+isinθ
1+a/1-a
=(1+cosθ+isinθ)/(1-cosθ-isinθ)
=(1+cosθ+isinθ)(1-cosθ+isinθ)/(1-cosθ+isinθ)(1-cosθ-isinθ)
=(1+cosθ+isinθ-cosθ-cos²θ-isinθcosθ+isinθ+isinθcosθ+i²sin²θ)/{(1-cosθ)²-(isinθ)²}
=(1+2isinθ-cos²θ-sin²θ)/{1-2cosθ+cos²θ-(i²sin²θ)}
={1+2isinθ-(sin²θ+cos²θ)}/(1-2cosθ+cos²θ+sin²θ)
=(1+2isinθ-1)/(1-2cosθ+1)
=2isinθ/(2-2cosθ)
=isinθ/(1-cosθ)
=i(2sinθ/2cosθ/2)/(2sin²θ/2)
=(icosθ/2)/(sinθ/2)
=icotθ/2
Hole it helps you if any problem.u can ask
A=cosθ+isinθ
1+a/1-a
=(1+cosθ+isinθ)/(1-cosθ-isinθ)
=(1+cosθ+isinθ)(1-cosθ+isinθ)/(1-cosθ+isinθ)(1-cosθ-isinθ)
=(1+cosθ+isinθ-cosθ-cos²θ-isinθcosθ+isinθ+isinθcosθ+i²sin²θ)/{(1-cosθ)²-(isinθ)²}
=(1+2isinθ-cos²θ-sin²θ)/{1-2cosθ+cos²θ-(i²sin²θ)}
={1+2isinθ-(sin²θ+cos²θ)}/(1-2cosθ+cos²θ+sin²θ)
=(1+2isinθ-1)/(1-2cosθ+1)
=2isinθ/(2-2cosθ)
=isinθ/(1-cosθ)
=i(2sinθ/2cosθ/2)/(2sin²θ/2)
=(icosθ/2)/(sinθ/2)
=icotθ/2
Hole it helps you if any problem.u can ask
Incredible29:
thnx a lot
u r welcome
actually i was typing this before u ased to solve it
oh :))
anything else
Answered by
8
Heya Friend !!
Please view the solution in the attachment .
hope this helps :)
thank you :)
Please view the solution in the attachment .
hope this helps :)
thank you :)
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welcome :)
and ur handwriting is so neat n mindblowing
Thank you so much...
:)
great handwriting bro i never write maths so neatly
I'm sis ^_^
oh i didn't saw that i juat saw dp
sorry
It's ok..xD
ok bye and sorry angaun sis
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