Class 11MathematicsTrigonometrical Ratios, Functions and IdentitiesTrigonometrical ratios of sum and difference of two and three angles
MediumProve that $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$.
(N/A) $L.H.S. = \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}$
$= \frac{(\cos 4x + \cos 2x) + \cos 3x}{(\sin 4x + \sin 2x) + \sin 3x}$
$= \frac{2 \cos(\frac{4x + 2x}{2}) \cos(\frac{4x - 2x}{2}) + \cos 3x}{2 \sin(\frac{4x + 2x}{2}) \cos(\frac{4x - 2x}{2}) + \sin 3x}$
Using the identities $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$ and $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$= \frac{2 \cos 3x \cos x + \cos 3x}{2 \sin 3x \cos x + \sin 3x}$
$= \frac{\cos 3x(2 \cos x + 1)}{\sin 3x(2 \cos x + 1)}$
$= \frac{\cos 3x}{\sin 3x} = \cot 3x = R.H.S.$
$= \frac{(\cos 4x + \cos 2x) + \cos 3x}{(\sin 4x + \sin 2x) + \sin 3x}$
$= \frac{2 \cos(\frac{4x + 2x}{2}) \cos(\frac{4x - 2x}{2}) + \cos 3x}{2 \sin(\frac{4x + 2x}{2}) \cos(\frac{4x - 2x}{2}) + \sin 3x}$
Using the identities $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$ and $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$= \frac{2 \cos 3x \cos x + \cos 3x}{2 \sin 3x \cos x + \sin 3x}$
$= \frac{\cos 3x(2 \cos x + 1)}{\sin 3x(2 \cos x + 1)}$
$= \frac{\cos 3x}{\sin 3x} = \cot 3x = R.H.S.$
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