Math, asked by Ashwati, 1 year ago

please answer this question.

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TahaTasneem: excuse me, is it a+b, or a+ib?

Ashwati: that's a+ib

Ankit1234: because very less user are online at this time

Ashwati: Sorry guys that was a wrong ques but this one is correct . Plzz ans this

Ashwati: i stand corrected

ankit321: I think its a+ib......bt the ques is wrong....

Ashwati: #ankit321 the ques is correct

ankit321: Oh!!!! Srryyy......I think the ques is wrong.......

Answers

Answered by ARoy

3/(2+cosθ+isinθ)=a+ib
or, 3(2+cosθ-isinθ)/(2+cosθ+isinθ)(2+cosθ-isinθ)=a+ib
or, 3(2+cosθ-isinθ)/{(2+cosθ)²-(isinθ)²}=a+ib
or, 3(2+cosθ-isinθ)/(4+4cosθ+cos²θ+sin²θ)=a+ib
or, 3(2+cosθ-isinθ)/(5+4cosθ)=a+ib [∵, sin²θ+cos²θ=1]
or, a+ib=3(2+cosθ)/(5+4cosθ)+i(-3sinθ)/(5+4cosθ)
Equating botth sides,
a=3(2+cosθ)/(5+4cosθ), b=-3sinθ/(5+4cosθ)
∴, LHS
=a²+b²={3(2+cosθ)/(5+4cosθ)}²+{-3sinθ/(5+4cosθ)}²
=9(4+4cosθ+cos²θ)/(5+4cosθ)²+9sin²θ/(5+4cosθ)²
=9(4+4cosθ+cos²θ+sin²θ)/(5+4cosθ)²
=9(5+4cosθ)/(5+4cosθ)²
=9/(5+4cosθ)
RHS
=4a-3
=4{3(2+cosθ)/(5+4cosθ)}-3
=12(2+cosθ)/(5+4cosθ)-3
=(24+12cosθ-15-12cosθ)/(5+4cosθ)
=9/(5+4cosθ)
∴, LHS=RHS (Proved)

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