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` ( 1 ) / ( 1 - cos theta + 2 i sin theta ) ` को ` a + ib ` के रूप में व्यक्त कीजिये | |
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Answer» ` ( 1 ) /( 1 - cos theta + 2 isin theta ) ` ` = ( 1 ) /( 1 - cos theta + 2i sin theta )* (( 1 - cos theta - 2i sin theta )) /( ( 1 - cos theta -2i sin theta)) ` ` = ( 1 - cos theta - 2i sin theta ) /( ( 1 - cos theta ) ^ 2 + 4 sin ^ 2 theta ) ` ` ( 2 sin ^ 2 "" ( theta ) /(2 ) - 2i. 2 sin "" ( theta ) /(2) cos "" ( theta ) /(2 ) )/( ( 2 sin ^ 2 "" ( theta ) /(2) ) ^ 2 + 4 ( 2 sin "" ( theta ) /(2) cos "" ( theta ) /(2) )^ 2 ) = ( 2 sin ^ 2 "" ( theta ) / ( 2) ( 1 - 2i cot""( theta ) / (2)) ) / ( 2 sin^ 2 "" ( theta ) /(2) ( 2 sin ^ 2 "" ( theta ) /(2) + 8 cos^ 2 "" ( theta ) /(2)) ) ` ` = ( 1 -2i cot"" ( theta ) / ( 2)) /( 2 ( sin ^2 "" ( theta ) /(2) + cos ^ 2 "" ( theta ) / (2) ) + 6 cos^ 2 ""( theta ) / ( 2 )) = ( 1 - 2 i cot"" ( theta ) /(2)) /( 2 + 3 ( 1 + cos theta )) ` ` = ( 1 - 2i cot"" ( theta ) / ( 2 ) ) /( 5 + 3 cos theta ) = ( 1 ) /( 5 + 3 cos theta ) - i ( 2 cot"" ( theta ) /(2))/( 5 + 3 cos theta ) ` |
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