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Prove that,
$\frac{ \cos4x + \cos3x + \cos2x}{ \sin4x + \sin3x + \sin2x } =cot3x$

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Answer 1

Given:-

L.H.S-$\frac{ \cos4x + \cos3x + \cos2x}{ \sin4x + \sin3x + \sin2x }$

R.H.S-$cot3x$

To Prove:-

• L.H.S=R.H.S

Solution:-

Solving L.H.S

$\frac{ \cos4x + \cos3x + \cos2x}{ \sin4x + \sin3x + \sin2x }$

$= > \frac{( \cos4x + \cos2x )+ \cos3x}{ (\sin4x + \sin2x )+ \sin3x}$

we know that :

$\sf \: \cos(x) + \cos(y) = 2 \cos( \frac{x + y}{2} ) \cos( \frac{x - y}{2} )$

and

$\sf\sin(x) + \sin(y) = 2 \sin( \frac{x + y}{2} ) \cos( \frac{x - y}{2} )$

Now, we will use these formula to solving L.H.S

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

$= > \frac{2 \cos( \frac{6x}{2} ) \cos( \frac{2x}{2} ) + \cos(3x) }{2 \sin( \frac{6x}{2} ) \cos( \frac{2x}{2} ) + \sin(3x) }$

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

Now, taking cos3x common in numerator and taking sin3x common in denominator

$= > \frac{ \cos3x(2 \cos(x) + 1) }{ \sin3x (2 \sin(x) + 1) }$

$= > \frac{ \cos3x \times ( 1) }{ \sin3x \times ( 1) }$

we know that

$\sf \frac{ \cos(x) }{ \sin(x) } = \cot(x)$

So,

$= > \cot(3x)$

as it equals to R.H.S

$\sf\therefore \: L.H.S=R.H.S$