Question

Prove that

$cotx \: cot2x - cot2x \: cot3x - \: cot3x \: cotx \: = 1$
​

Answer 1

Answer:

taking L.H.S

cot x cot 2x -cot 2x cot 3x -cot 3x cotx

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

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

Answer 2

LHS = (cotx.cot2x) - (cot2x.cot3x) - (cot3x.cotx )

We know,

2x + x = 3x

taking cot on both sides,

cot(x + 2x) = cot3x

We know,

$\huge\boxed{\sf cot(A + B) = \dfrac{(cotA.cotB-1)}{(cotA+CotB)}}$

Applying the values to the equation,

$\implies \sf cot(3x) = \dfrac{(cotx.cot2x-1)}{(cotx+Cot2x)}$

$\implies\sf cotx.cot2x -1 = cot3x(cotx+cot2x)$

$\implies\sf cotx.cot2x -1 = cot3x.cotx + cot3x.cot2x$

$\bold{\implies\sf cotxcot2x - cot2x.cot3x - cot3x.cotx = 1}$

$\huge\boxed{\bold{\rm Hence\:proved}}$