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Trigonometry sum

I need some help in proving the statement:

$\frac{1}{\cos x \cos 2x} + \frac{1}{\cos 2x \cos 3x} + ... + \frac{1}{\cos nx \cos (n+1)x} =$ $\frac{\tan (n+1)x}{\sin x}$

I don't think this is true. For n=1, for example, this would suggest that

$<br /> \frac 1 {\cos(x) \cos(2x)} = \frac {\tan(2x)} {\sin (x)}<br />$

But the right hand side is:

$<br /> \frac {\tan (2x)}{\sin (x)} =\frac {\sin (2x)}{\sin x \cos (2x)} = \frac {2 \sin (x) \cos (x)}{\sin (x) \cos (2x)} = \frac {2 \cos (x)}{\cos (2x)}<br />$

which does not equal the left hand side.

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