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Answer 1

Question:-

Prove :

$\frac{ \sin8x \cos \: x - \cos3x \sin6x }{ \cos2x \cos \: x - \sin3x \sin4x } = \tan2x$

Solution:-

To Prove:-

$\frac{ \sin8x \cos \: x - \ \sin6x }{ \cos2x \cos \: x - \sin3x \sin4x } = \tan2x$

Procedure:-

Take LHS

Multiply numerator and denominator with 2

$\frac{(2 \sin8x \cos \: x - 2 \sin6x \cos3x)}{(2 \cos2x \cos \: x - 2 \sin3x \sin4x)}$

Apply product to sum formula

$\frac{ (\sin9x + \sin7x) - ( \sin9x + \sin3x)}{( \cos3x + \cos \: x) - ( - \cos \: x - \cos7x)}$

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

Apply sum to product formulae

$\frac{(2 \ \sin2x)}{(2 \cos5x \cos2x)}$

$\tan2x =RHS$

Hence, proved