Question

Prove that $cotx.cot2x-cot2x.cot3x-cotx.cot3x=1.$

Answer 1

$L.H.S.=cotx.cot2x-cot2x.cot3x-cotx.cot3x$

On taking out $cot3x$ as common factor we get,

$=cotx.cot2x-cot3x.(cot2x+cotx)$……………$\left(i\right)$

$cot3x$ can be expressed as $cot\left(2x+x\right)$

From the identity we know that,

$cot(A+B)=\frac{(cotA.cotB-1)}{cotA-cotB}$

Thus, $cot(2x+x)=\frac{(cot2x.cotx–1)}{(cot2x+cotx)}$

Equation $\left(i\right)$ becomes,

$=cotx.cot2x-\left(\frac{cot2x.cotx–1}{cot2x+cotx}\right)(cot2x+cotx)$

$=cotx.cot2x-cot2x.cotx+1$

$=1$.

$=R.H.S.$

Thus, it is proved that, $cotx.cot2x-cot2x.cot3x-cotx.cot3x=1.$