If tan^2 alpha tan^2 beta + tan^2 beta tan^2 | vedclass

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(C) Let $x = 	an^2 \alpha$,$y = 	an^2 \beta$,and $z = 	an^2 \gamma$.The given equation is $xy + yz + zx + 2xyz = 1$.We want to find the value of $S

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$1$

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(C) Let $x = 	an^2 \alpha$,$y = 	an^2 \beta$,and $z = 	an^2 \gamma$.The given equation is $xy + yz + zx + 2xyz = 1$.We want to find the value of $S = \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.Since $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	heta}$,we have $S = \frac{x}{1+x} + \frac{y}{1+y} + \frac{z}{1+z}$.$S = \frac{x(1+y)(1+z) + y(1+x)(1+z) + z(1+x)(1+y)}{(1+x)(1+y)(1+z)}$.Expanding the numerator: $x(1+y+z+yz) + y(1+x+z+xz) + z(1+x+y+xy) = x+xy+xz+xyz + y+yx+yz+xyz + z+zx+zy+xyz = (x+y+z) + 2(xy+yz+zx) + 3xyz$.Using the identity $(1+x)(1+y)(1+z) = 1 + (x+y+z) + (xy+yz+zx) + xyz$,and substituting $xy+yz+zx = 1 - 2xyz$,we get:$S = \frac{(x+y+z) + 2(1-2xyz) + 3xyz}{1 + (x+y+z) + (1-2xyz) + xyz} = \frac{x+y+z + 2 - 4xyz + 3xyz}{2 + x+y+z - xyz} = \frac{x+y+z + 2 - xyz}{2 + x+y+z - xyz} = 1$.

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 the value of $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$ is

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