Prove that cos 4x - cos 3x - cos 2xsin 4x - sin 3x - sin 2x - cot 3x

cos 4x + cos 3x + #10;cos 2x/sin 4x + sin 3x + sin 2x = cos 3x + 2cos 3xcos #10;x/sin 3x + 2sin 3xcos x #10; #10; = cos 3x( 1 #10;+ 2cos x)/ ...

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Question

Prove that $\frac{c o s 4 x + c o s 3 x + c o s 2 x}{s i n 4 x + s i n 3 x + s i n 2 x} = c o t 3 x$

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\frac{cos 4 x + cos 3 x + cos 2 x}{sin 4 x + sin 3 x + sin 2 x}

Prove the following:
$\frac{cos 4x+ cos 3x + cos 2x}{sin 4x + sin 3x + sin 2x} = cot 3 x$

\frac{s i n 2 x - s i n 3 x - s i n 4 x}{c o s 2 x - c o s 3 x + c o s 4 x} = t a n 3 x .

y = \frac{cos x + cos 2 x + cos 3 x + cos 4 x + cos 5 x + cos 6 x + cos 7 x}{sin x + sin 2 x + sin 3 x + sin 4 x + sin 5 x + sin 6 x + sin 7 x},

\cos x + \cos 2 x + \cos 3 x = 0

\cos x + \cos 3 x - \cos 2 x = 0

\sin x + \sin 5 x = \sin 3 x

\cos x \cos 2 x \cos 3 x = \frac{1}{4}

\cos x + \sin x = \cos 2 x + \sin 2 x

\sin x + \sin 2 x + \sin 3 = 0

\sin x + \sin 2 x + \sin 3 x + \sin 4 x = 0

\sin 3 x - \sin x = 4 \cos^{2} x - 2

\sin 2 x - \sin 4 x + \sin 6 x = 0

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