one more identity to prove

**nightrider456** · January 18th 2010, 05:59 PM

Thanks for having a look.

$<br /> \frac{sin2x}{cosx}+\frac{cos2x}{sinx}=cscx<br />$

**Prove It** · January 18th 2010, 06:09 PM

Originally Posted by nightrider456

Thanks for having a look.

$<br /> \frac{sin2x}{cosx}+\frac{cos2x}{sinx}=cscx<br />$

$\frac{\sin{2x}}{\cos{x}} + \frac{\cos{2x}}{\sin{x}} = \frac{2\sin{x}\cos{x}}{\cos{x}} + \frac{\cos^2{x} - \sin^2{x}}{\sin{x}}$

$= 2\sin{x} + \frac{\cos^2{x} - \sin^2{x}}{\sin{x}}$

$= \frac{2\sin^2{x} + \cos^2{x} - \sin^2{x}}{\sin{x}}$

$= \frac{\sin^2{x} + \cos^2{x}}{\sin{x}}$

$= \frac{1}{\sin{x}}$

$= \csc{x}$ .

**Krizalid** · January 19th 2010, 06:16 AM

$\frac{\sin 2x}{\cos x}+\frac{\cos 2x}{\sin x}=\frac{\cos (2x-x)}{\sin (x)\cos (x)}=\csc x.$

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