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यदि ${\tan ^2}\alpha \;{\tan ^2}\beta + {\tan ^2}\beta \;{\tan ^2}\gamma + {\tan ^2}\gamma \;{\tan ^2}\alpha + 2{\tan ^2}\alpha \;{\tan ^2}\beta \;{\tan ^2}\gamma = 1,$ तब
${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma $ का मान है
$0$
$-1$
$1$
इनमें से कोई नहीं
Solution
(c) ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma $
$ = \frac{{{{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }} + \frac{{{{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }} + \frac{{{{\tan }^2}\gamma }}{{1 + {{\tan }^2}\gamma }}$
$ = \frac{x}{{1 + x}} + \frac{y}{{1 + y}} + \frac{z}{{1 + z}}$ $(x = {\tan ^2}\alpha ,\,y = {\tan ^2}\beta ,\,z = {\tan ^2}\gamma )$
$ = \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$ $( \because xy + yz + zx + 2xyz = 1)$
