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

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(C) Given: $	an^2 \alpha 	an^2 \beta + 	an^2 \beta 	an^2 \gamma + 	an^2 \gamma 	an^2 \alpha + 2	an^2 \alpha 	an^2 \beta 	an^2 \gamma = 1$.Let

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(C) Given: $	an^2 \alpha 	an^2 \beta + 	an^2 \beta 	an^2 \gamma + 	an^2 \gamma 	an^2 \alpha + 2	an^2 \alpha 	an^2 \beta 	an^2 \gamma = 1$.Let $x = 	an^2 \alpha$,$y = 	an^2 \beta$,and $z = 	an^2 \gamma$.Then the given equation is $xy + yz + zx + 2xyz = 1$.We need to find $S = \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.Using $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	heta}$,we have:$S = \frac{x}{1+x} + \frac{y}{1+y} + \frac{z}{1+z}$.$S = \frac{x(1+y)(1+z) + y(1+x)(1+z) + z(1+x)(1+y)}{(1+x)(1+y)(1+z)}$.$S = \frac{x(1+y+z+yz) + y(1+x+z+xz) + z(1+x+y+xy)}{(1+x)(1+y)(1+z)}$.$S = \frac{x+xy+xz+xyz + y+xy+yz+xyz + z+xz+yz+xyz}{(1+x)(1+y)(1+z)}$.$S = \frac{(x+y+z) + 2(xy+yz+zx) + 3xyz}{(1+x)(1+y)(1+z)}$.Since $xy+yz+zx+2xyz = 1$,we can substitute $xy+yz+zx = 1-2xyz$.$S = \frac{(x+y+z) + 2(1-2xyz) + 3xyz}{1 + (x+y+z) + (xy+yz+zx) + xyz}$.$S = \frac{x+y+z + 2 - 4xyz + 3xyz}{1 + (x+y+z) + (1-2xyz) + xyz} = \frac{x+y+z + 2 - xyz}{x+y+z + 2 - xyz} = 1$.

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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