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

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(c) ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma $

$ = \frac{{{{	an }^2}\alpha }}{{1 + {{	an }^2}\alpha }} + \frac{{{{	an }^2}\beta }}{{1

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(c) ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma $

$ = \frac{{{{	an }^2}\alpha }}{{1 + {{	an }^2}\alpha }} + \frac{{{{	an }^2}\beta }}{{1 + {{	an }^2}\beta }} + \frac{{{{	an }^2}\gamma }}{{1 + {{	an }^2}\gamma }}$

$ = \frac{x}{{1 + x}} + \frac{y}{{1 + y}} + \frac{z}{{1 + z}}$ $(x = {	an ^2}\alpha ,\,y = {	an ^2}\beta ,\,z = {	an ^2}\gamma )$

$ = \frac{{(x + y + z) + (xy + yz + zx + 2xyz) + xy + yz + zx + xyz}}{{(1 + x)(1 + y)(1 + z)}}$ 

$ = \frac{{1 + x + y + z + xy + yz + zx + xyz}}{{(1 + x)(1 + y)(1 + z)}} = 1$       $( \because xy + yz + zx + 2xyz = 1)$

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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