Question

Hey!___96 points___
plss plsss solve it...........

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Answer 1

Step-by-step explanation:

Taking LHS :

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

[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