`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`

Explanation

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))`

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