If `cot^(3)alpha+cot^(2)alpha+cot alpha=1` then which of the following is/are correct

Options

A. `cos2alpha.tanalpha=-1`
B. `cos 2alpha.tan alpha=1`
C. `cos 2alpha-tan 2alpha=1`
D. `cos 2alpha-tan 2alpha=1`.

Answer

A, C

Explanation

`cot^(3) alpha + cot^(2) alpha + cot alpha =1 `
`rArr " " tan ^(3) alpha - tan^(2) alpha - tan alpha - 1=0 `
`rArr " " tan alpha ( tan ^(2) alpha - 1) = 1 + tan^(2) alpha `
`rArr " " tan alpha = - ((1+ tan^(2) alpha)/(1-tan^(2)alpha))`
`rArr " " cos 2alpha * tan alpha = -1`
Now `cos 2alpha - tan 2alpha = - (1)/(tan alpha ) - ( 2 tan alpha ) /(1-tan^(2) alpha)`
`" " = - ((1-tan^(2)alpha + 2 tan ^(2) alpha)/( tan alpha (1-tan^(2)alpha)))`
`" " = -(1 + tan^(2)alpha)/(tanalpha (1-tan^(2) alpha ))`
`" " =- (1)/(tan alpha cos 2 alpha)`
`" " =1 `