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  • Prove the following (cos9x-cos5x)/ (sin17x -sin3x)= - sin2x/ cos10x

Q (16) Prove the following

\small \frac{\cos 9x - \cos 5x}{\sin17x - \sin3x} = -\frac{\sin2x}{\cos10x}

Answers (1)

As we know that

\\ \cos x - \cos y = -2\sin\frac{x+y}{2}\sin\frac{x-y}{2 }\\ \\ \cos 9x - \cos 5x = -2\sin 7x \sin2x \\ \\ \sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2 }\\ \\ \sin 17x - \sin 3x = 2\cos10x \sin7x\\ \\ \frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} =\frac{-2\sin 7x \sin2x}{2\cos10x \sin7x} = -\frac{\sin 2x}{\cos10x}
                                                                                                        R.H.S.

Posted by

Safeer PP

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