If `sin(x +3alpha)=3 sin (alpha-x)`, then

Options

A. `tan x = tan alpha`
B. `tan x = tan^(2)alpha`
C. `tanx=tan^(3)alpha`
D. `tanx=3tanalpha`

Answer

C

Explanation

Given, `(sin(x+3alpha))/(sin(alpha-x))=3`
On applying componendo and dividendo rule, we get
`(sin(x+3alpha)+sin(alpha-x))/(sin(x+3alpha)-sin(alpha-x))=(3+1)/(3-1)`
`implies (2sin2alphacos(alpha+x))/(2cos2alphasin(alpha+x))=2`
`implies (tan2alpha)/(tan(alpha+x))=2`
`implies (2tanalpha)/(1-tan^(2)alpha)xx((1-tanalphatanx))/((tanalpha+tanx))=2`
`implies tanalpha-tan^(2)alphatanx=tanalpha+tanx-tan^(3)alpha-tan^(2)alphatanx`
`impliestanx=tan^(3)alpha`