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If, a=cosα+isinα,b=cosβ+isinβ and c=cosγ+isinγ find the value of, abccab.
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Solution:

Given:

a=cosα+isinαb=cosβ+isinβ&c=cosγ+isinγ

Now,

abc=ab(c)1

=(cosα+isinα)(cosβ+isinβ)(cosγ+isinγ)1

=(cosα+isinα)(cosβ+isinβ)(cosγisinγ)

abc=cos(α+βγ)+isin(α+βγ)

also,cab=[[abc]1]

=[cos(α+βγ)+isin(α+βγ)]1

cab=cos(α+βγ)isin(α+βγ)

(1)(2) then,

abccab=cos(α+βγ)+isin(α+βγ)cos(α+βγ)+isin(α+βγ) abccab=2isin(α+βγ)

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