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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove : $\large\frac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\frac{\sin 2x}{\cos 10x}$

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Toolbox:
  • $\cos A-\cos B=-2\sin \large\frac{A+B}{2}$$\sin \large\frac{A-B}{2}$
  • $\sin A-\sin B=2\cos \large\frac{A+B}{2}$$\sin \large\frac{A-B}{2}$
L.H.S
$\large\frac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}$
$\Rightarrow \large\frac{-2\sin \Large\frac{9x+5x}{2}.\sin \Large\frac{9x-5x}{2}}{2\cos \Large\frac{17x+3x}{2}\sin \Large\frac{17x-3x}{2}}$
$\Rightarrow \large\frac{-\sin 7x.\sin 2x}{\cos 10x.\sin 7x}$
$\Rightarrow -\large\frac{\sin 2x}{\cos 10x}$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 
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