prove the following

Q(22) \small \cot x \cot2x - \cot2x\cot3x - \cot3x\cot x =1

Answers (1)

cot x cot2x - cot3x(cot2x - cotx)
Now we can write cot3x = cot(2x + x)

and we know that

cot(a+b) = \frac{\cot a \cot b - 1}{\cot a + \cot b}
 So,    
             cotx\ cot2x-\frac{\cot 2x \cot x - 1}{\cot 2x + \cot x}(cot2x+cotx)   
           =   cotx cot2x - (cot2xcotx -1)
           =  cotx cot2x - cot2xcotx +1
            = 1  = R.H.S.

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