The value of tan `alpha +2tan(2alpha)+4tan(4alpha)+...+2^(n-1)tan(2^(n-1)alpha)+2^ncot(2^nalpha)` is

Options

A. `cos (2^(n)alpha)`
B. `2^(n) tan (2^(n) alpha)`
C. 0
D. `cot alpha`

Answer

D

Explanation

We have,
`cot theta -tantheta=2cot 2 theta`
`thereforetan alpha+2 tan2 alpha+4tan4 alpha+...+tan(2^(n-1)alpha)+2^(n)cos(2^(n)alpha)`
`=-{cotalpha-tanalpha)=2tan2 alpha-4tan4 alpha...-2^(n-1)tan(2^(n-1)alpha)-2^(n)cot(2^(n)alpha)}+cot alpha`
`={-(2 cot2 alpha-2tan2 alpha)-4tan 4 alpha... -2^(n-1)tan(2^(n-1)alpha)-2^(n)cot(2^(n)alpha)}+cot alpha`
`={-(2^(2) cot2^(2) alpha-2^(2)tan2 alpha)-2^(3)tan 2^(3) alpha... -2^(n-1)tan(2^(n-1)alpha)-2^(n)cot(2^(n)alpha)}+cot alpha`
`=-{(2^(3)cot2^(3)alpha-2^(3)tan2^(3)alpha)-2^(4)tan2^(4)alpha...-^(n-1)tan(2^(n-1)alpha)-2^(n)cot(2^(n)alpha)}+cot alpha`
`=-{2^(n-1)cot(2^(n-1)alpha)-2^(n-1)tan(2^(n-1)alpha)-2^(n)cot(2^(n)alpha)}+cotalpha`
`=-{2^(n-1)xx2cot(2^(n)alpha)-2^(n)cot(2^(n)alpha)}+cot alpha=cot alpha`