If `tan^2 alpha tan^2 beta + tan^2 beta tan^2 gamma + tan^2 gamma tan^2 alpha + 2 tan^2 alpha tan^2 beta tan^2 gamma = 1` then ` sin^2 alpha + sin^2 beta + sin^2 gamma` =

Options

A. 0
B. 1
C. 1
D. none of these

Answer

C

Explanation

`sin^(2)alpha+sin^(2)beta+sin^(2)gamma`
`=(tan^(2)alpha)/(1+tan^(2)alpha)+(tan^(2)beta)/(1+tan^(2)beta)+(tan^(2)gamma)/(1+tan^(2)gamma)`
`=(x)/(1+x)+(y)/(1+y)+(z)/(1+z)` [where `x=tan^(2)alpha,gamma=tan^(2)beta, z= tan^(2)gamma`]
`=((x+y+z)+(xy+yz+zx+2xyz)+xy+zx+xyz)/((1+x)(1+y)(1+z))`
`=(1+x+y+z+xy+yz+zx+xyz)/((1+x)(1+y)(1+z))=1`
`[because xy+yz+zx+2xyz+2xyz=1]`