Question

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

Answer 1

Answer:

cot x cot 2x - cot 2x cot 3x - cot 3x cot 4x 

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

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

[ as cot(A+B)=cotAcotBcotAcotB−1]

= cot x cot 2x-(cotx+cot2xcot2xcotx−1)(cot 2x+ cot x)

= cot x cot 2x- (cot 2x cot x-1)

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

= 1

Hence, proved

Answer 2

Step-by-step explanation:

Solution :

We have,

$\: \: \: \: \: \: \: \cot3x = \cot(2x + x)$

$= > \cot3x = \frac{ \cot2x \cot \: x - 1 }{ \cot2x + \cot \: x }$

$= > \cot3x \cot2x + \cot3x \cot \: x = \cot2x \cot \: x - 1$

$= > \cot2x \cot \: x - \cot3x \cot2x - \cot3x \cot \: x = 1.$

Hence :

Proved.