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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: tanx..cot x..
Forum Index -> Trignometry like the article? email it to a friend.  
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mathwiz (46)

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1/(tan 3x-tan x)  -  1/(cot 3x-cotx)=?
    
uday_zingtudor (931)

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Write cot3x as 1/tan3x and cotx as 1/tanx

Now the expression reduces to the form 1+tan3xtanx/tan3x-tanx

And that 's equal to 1/tan2x = cot2x

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mathwiz (46)

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thanks..got it
 
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pantpranav (341)

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There's one more way to solve it:


tan 3x - tan x = sin (3x- x) / (cos 3x cos x)


& cot 3x - cot x = sin(x - 3x) / (sin 3x sin x)


Putting in the given expression:


= cos 3x cos x / sin 2x - sin 3x sin x / sin (-2x)


= (cos 3x cos x + sin 3x sin x) / sin 2x


= cos 2x / sin 2x


= cot 2x.


This gives the same result.

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