Question

prove that sin 8x cosx - cos 3x sin 6x÷ cos2xcosx- sin3x sin4x =tan 2x

Answer 1

Answer:

Step-by-step explanation:

L.H.S.

$\frac{\sin 8x \cos x -\sin 6x \cos 3x}{\cos 2x\cos x -\sin 3x \sin 4x}\\\\=\frac{2\sin 8x \cos x -2\sin 6x \cos 3x}{2\cos 2x\cos x -2\sin 3x \sin 4x}\\\\=\frac{\sin (8x+x)+\sin (8x-x)-(\sin (6x+3x)+\sin (6x-3x))}{\cos (2x+x)+\cos (2x-x)-(\cos (4x-3x)-\cos (4x+3x))}\\\\=\frac{\sin 9x+\sin 7x-\sin 9x-\sin 3x}{\cos 3x+\cos x-\cos x+\cos 7x}\\\\=\frac{\sin 7x-\sin 3x}{\cos 7x+\cos 3x}\\\\=\frac{2\sin (\frac{7x-3x}{2})\cos \frac{(\frac{7x-3x}{2})}{2} }{2\cos (\frac{7x+3x}{2})\cos (\frac{7x-3x}{2})}\\\\=\frac{\sin 2x\cos 5x}{\cos 2x\cos 5x}\\\\=\frac{\sin 2x}{\cos 2x}\\\\=\tan 2x$

= R.H.S

hence proved.

Answer 2

Answer:

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Step-by-step explanation:

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