Question

Prove that
(sin3xcosx-sin6xcos3x)/(cos2xcosx-sin4xsin3x) = tan2x
pls help i need it urgently
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Answer 1

Correct question:-

Prove that

$\\\frac {(sin8xcosx-sin6xcos3x)}{(cos2xcosx-sin4xsin3x)}$ = tan2x

Answer:-

LHS =

$\\\frac {(sin8xcosx-sin6xcos3x)}{(cos2xcosx-sin4xsin3x)}$

By multiplying numerator and denominator by 2,

$\\=\frac {(2sin8xcosx-2sin6xcos3x)}{(2cos2xcosx-2sin4xsin3x)}$

$By\:\:using\: \:formula\:\:(Product\: \:to \:\:sum) ,$

$\boxed{2sinAcosB = sin(A+B) + sin(A-B) }$

$\boxed{-2sinAsinB = cos(A+B) -cos(A-B) }$

$\boxed{2cosAcosB = cos(A+B) + cos(A-B)}$

$\\=\frac {[sin(8x+x)+sin(8x-x)]- [sin(6x+3x)+sin(6x-3x)]}{[cos(2x+x)+cos(2x-x)]+[cos(4x+3x) -cos(4x-3x)]}$

$\\=\frac{[sin9x+sin7x]- [sin9x+sin3x]}{[cos3x+cosx]+[cos7x-cosx]}$

$\\=\frac {sin7x-sin3x}{cos3x+cos7x}$

By using formula (Sum to product) ,

$\\=\frac {2×cos[(7x+3x)/2]×sin[(7x-3x)/2]}{2×cos[(7x+3x)/2]×cos[(7x-3x)/2]}$

$\\=\frac {2×cos5x×sin2x}{2×cos5x×cos2x}$

$\\=\frac {sin2x}{cos2x}$

= tan2x

= RHS

HENCE PROVED