sin2theta=2tantheta/1+tan^2theta
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Hi ,
RHS = 2tanA /( 1 + tan²A )
= ( 2sinA/cosA )/( 1 + sin²A/cos²A )
= ( 2sinA/cosA )/[( cos²A + sin²A )/cos²A ]
= ( 2sinAcosA )/( cos²A + sin²A )
= 2sinAcosA
[ cos²A + sin²A = 1 ]
= sin 2A
[ since , 2sinAcosA = sin2A ]
= LHS
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RHS = 2tanA /( 1 + tan²A )
= ( 2sinA/cosA )/( 1 + sin²A/cos²A )
= ( 2sinA/cosA )/[( cos²A + sin²A )/cos²A ]
= ( 2sinAcosA )/( cos²A + sin²A )
= 2sinAcosA
[ cos²A + sin²A = 1 ]
= sin 2A
[ since , 2sinAcosA = sin2A ]
= LHS
** If it really helped u.
invite your friends to brainly **
: )
: )
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