# Prove the given identity [duplicate]

$\dfrac{2\sin x}{\cos 3x} + \dfrac{2\sin 3x}{\cos 9x} + \dfrac{2\sin 9x}{\cos 27x} = \tan 27 x- \tan x$.

I did not get even to start with which formula. I tried using multiple angle identities but did not find any suitable place to use those. So please help.

## marked as duplicate by lab bhattacharjee trigonometry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 6 '16 at 12:33

• The question already exists – Archis Welankar Jan 6 '16 at 12:11
• @ Archis where does it exist? I haven't posted it before – Ger Wyn Jan 6 '16 at 12:15
• There are many people who ask questions please see tags. – Archis Welankar Jan 6 '16 at 12:21

See if this leads you anywhere. I've included an example for your first term:

$\dfrac{2\sin x}{\cos3x} = \dfrac{\dfrac{2\sin x\cos x}{\cos x}}{\cos3x} = \dfrac{2\sin x\cos x}{\cos x \times \cos3x} = \dfrac{2\sin x\cos x}{\dfrac{1}{2}(\cos4x + \cos2x)}$

One advantage of this is you may be able to begin condensing using the double angle formula. Also note that they're all in multiples of 3, i.e $x$ and $3x$, $3x$ and $9x$, etc. so a few substitutions may help.

EDIT: OP, I'd recommend you to take a look at the answers on this post when it was previously asked i.e here. Particularly Lab Bhatacharjee's post is simple and insightful, although I myself am not yet familiar with telescoping series (I am a sophomore in high school). Also, if you'd like an inductive approach, take a look at Leg's answer.