if `(tanalpha-i(sin ""(alpha)/(2)+cos ""(alpha)/(2)))/(1+2 i sin ""(alpha)/(2))` is purely imaginary then `alpha` is given by -

Options

A. `n pi +(pi)/(4)`
B. `n pi =(pi)/(4)`
C. `(2n+1) pi`
D. `2 n pi +(pi)/(4)`

Answer

A

Explanation

`([tan alpha -i(sin ""(alpha)/(2)+cos""(alpha)/(2))])/((1+2i sin ""(alpha)/(2))).((1-2i sin ""(alpha)/(2)))/((1-2isin""(alpha)/(2)))`
`([tan alpha-2.sin ""(alpha)/(2)(sin""(alpha)/(2)+cos""(alpha)/(2))]+i(2 tanalpha(sin""(alpha)/(2))-(sin""(alpha)/(2)+cos""(alpha)/(2))))/((1+4sin^(2)""(alpha)/(2)))`
`implies tan alpha - 2 sin ""(alpha)/(2).(sin""(alpha)/(2)+cos""(alpha)/(2))=0`
` tan alpha = 2 sin ((alpha)/(2)).(sin""(alpha)/(2)+cos""(alpha)/(2))`
`(cos""(alpha)/(2))/(cosalpha)=(sin ""(alpha)/(2)+cos""(pi)/(2))`
By solving `alpha = n pi +(pi)/(4)`