Question

`2tan^2 alpha tan^2 betatan^2gamma+tan^2 alpha tan^2beta+tan^2 betatan^2gamma+tan^2gamma tan^2 alpha=1` find the value of `sin^2 alpha+sin^2 beta +sin^2 gamma`

Answer 1

we know, `tan^2 alpha = sin^2alpha/cos^2alpha = sin^2alpha/(1-sin^2alpha)`
Let `sin^2alpha = x`
Then, `tan^2alpha = x/(1-x)`
Similarly, if `sin^2beta = y, then, tan^2beta = y/(1-y)`
Similarly, if `sin^2gamma= z, then, tan^2gamma = z/(1-z)`
Now, putting these values in the given equation,
`2(x/(1-x))(y/(1-y))(z/(1-z))+(x/(1-x))(y/(1-y))+(y/(1-y))(z/(1-z))+(z/(1-z))(x/(1-x)) = 1`
`=>(xy)/((1-x)(1-y))+(yz)/((1-y)(1-z))+(zx)/((1-z)(1-x)) = 1- (2xyz)/((1-x)(1-y)(1-z))`
`=>(xy-xyz+yz-xyz+zx-xyz)/((1-x)(1-y)(1-z)) = 1- (2xyz)/((1-x)(1-y)(1-z))`
`=>(x+y+z) - (2xyz)/((1-x)(1-y)(1-z)) = 1 - (2xyz)/((1-x)(1-y)(1-z))`
`=>x+y+z = 1`
`:. sin^2alpha+sin^2beta+sin^2gamma = 1`