Hey!___96 points___plss plsss solve it...........

1.	Hey!___96 points___plss plsss solve it...........
Answer» Step-by-step explanation: Taking LHS : $\frac{ \sin(2x) - \sin(3x) + \sin(4x) }{ \<klux>COS</klux>(2x) - \cos(3x) + \cos(4x) }$ [note that: $\frac{2x + 4x}{2} = 3x$ $= > \frac{( \sin(2x )+ \sin(4x) - \sin(3x) ) }{( \cos(2x) + \cos(4x) - \cos(3x) }$ ALSO we know that: sin(A+B) = 2sin((A+B)/2)cos((A-B)/2) =>cos(A+B) = 2 cos((A+B) /2)cos(A-B)/2) $= > \frac{2 \sin( \frac{2x + 4x}{2} ) \times \cos( \frac{2x - 4x}{2} ) - \sin(3x) }{2 \cos( \frac{2x + 4x}{2} ) \times \cos( \frac{2x - 4x}{2} ) - \cos(3x) }$ $= > \frac{2 \sin(3x) \times \cos( - x) - \sin(3x) }{2 \cos(3x) \times \cos( - x) - \cos(3x) }$ $= > \frac{ \sin(3x))(2 \cos(x) - 1)) }{ \cos(3x)((2 \cos(x) - 1 )) }$ *Now* *the* *same* *term* *will* *cancel* *out* *THEREFORE* : $= > \frac{ \sin(3x) }{ \cos(3x) }$ => *TAN3X* = *RHS* *Hence* *proved! HOPE* IT HELPS YOU

Hey!_96 points_plss plsss solve it...........

Answer»

Step-by-step explanation:

Taking LHS :

$\frac{ \sin(2x) - \sin(3x) + \sin(4x) }{ \<klux>COS</klux>(2x) - \cos(3x) + \cos(4x) }$

[note that:

$\frac{2x + 4x}{2} = 3x$

$= > \frac{( \sin(2x )+ \sin(4x) - \sin(3x) ) }{( \cos(2x) + \cos(4x) - \cos(3x) }$

ALSO we know that:

sin(A+B) = 2sin((A+B)/2)cos((A-B)/2)

=>cos(A+B) = 2 cos((A+B) /2)cos(A-B)/2)

$= > \frac{2 \sin( \frac{2x + 4x}{2} ) \times \cos( \frac{2x - 4x}{2} ) - \sin(3x) }{2 \cos( \frac{2x + 4x}{2} ) \times \cos( \frac{2x - 4x}{2} ) - \cos(3x) }$