Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

+1 vote
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asked Aug 9, 2017 in Mathematics by sforrest072 (157,439 points) 63 450 1293

2 Answers

+2 votes
answered Aug 10, 2017 by Rohit Singh (61,782 points) 36 143 463
selected Aug 10, 2017 by sforrest072
 
Best answer
L.H.S=cot x. cot 2x - cot 2x. cot3x - cot x. cot3x

=cot x. cot 2x - cot3x(cot 2x +cot x)

=cotx .cot 2x - cot (x+2x) .(cot 2x+ cot x)                       [we can write cot3x as cot(x+2x)]

=cotx.cot2x- {(cot x. cot 2x -1)/(cot x + cot 2x)}.(cot x + cot 2x)            [since cot(a+b)=(cot a. cot b -1)/(cot b+cot a)]

=cot x. cot 2x -cot x. cot 2x +1

=1

=R.H.S

+1 vote
answered Aug 10, 2017 by Rohit Singh (61,782 points) 36 143 463

Another method: 

Find out, what's the biggest angle here? It's "cot 3x"

cot (3x) = cot ( 2x+x )

Apply, cot (A+B) = (cot A.cot B -1)/ cot A + cot B

So,

cot (3x) = ( cot 2x.cot x - 1 ) / cot 2x + cot x

Now, cross multiply

cot 3x ( cot 2x + cot x ) = cot 2x.cot x - 1

cot 2x.cot 3x + cot x.cot 3x = cot 2x.cot x - 1

cot 2x.cot x - cot 2x.cot 3x - cot x.cot 3x = 1

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