If `(tan(alpha+beta+gamma))/("tan"(alpha-beta-gamma))=(tangamma)/(tanbeta),(beta!=gamma)` then `sin2alpha+s in2beta+s in2gamma=` 0 (b) 1 (c) 2 (d)

Options

A. 0
B. 1
C. 2
D. `1//2`

Answer

A

Explanation

`(tan (alpha+beta-gamma))/(tan(alpha-beta+gamma))=(tan gamma)/(tan beta)`
`rArr(sin(alpha+beta-gamma)cos(alpha-beta+gamma))/(sin(alpha-beta+gamma)cos(alpha+beta-gamma))=(sin gamma cos beta)/(sin beta cos gamma)`
Applying componendo and dividendo, we get
`(sin 2alpha)/(sin 2(beta-gamma))=(sin(gamma+beta))/(sin(gamma-beta))`
`rArr sin2(beta-gamma)sin(beta+gamma)+sin 2alpha sin(beta-gamma)=0`
`rArr sin(beta-gamma)(sin 2alpha+sin2beta+sin2gamma)=0`
`rArr sin2alpha+sin 2beta+sin 2y=0` (as `beta ne gamma`)