3.Trigonometrical Ratios, Functions and Identities
hard

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

A

$0$

B

$-1$

C

$1$

D

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$       $( \because xy + yz + zx + 2xyz = 1)$

Standard 11
Mathematics

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