Prove that cot A + cot (60 + A) –  cot (60 – A) = 3 cot 3A.

Solution :

We have,

L.H.S = cot A + cot (60 + A) –  cot (60 – A)

$\implies$ L.H.S = $1\over tan A$ + $1\over tan (60 + A)$ – $1\over tan (60 – A)$

$\implies$ L.H.S = $1\over tan A$+ $1 – \sqrt{3} tan A\over \sqrt{3} + tan A$ – $1 + \sqrt{3} tan A\over \sqrt{3} – tan A$

[ By using this formula, tan (A + B) = $tan A + tan B\over 1 – tan A tan B$  above ]

$\implies$ L.H.S = $1\over tan A$ – $8 tan A\over 3 –  tan^2 A$

$\implies$ L.H.S = $3 – 9 tan^2 A\over 3 tan A – tan^3 A$

L.H.S = 3($1 – 3 tan^2 A\over 3 tan A – tan^3 A$)

Since $3 tan A – tan^3 A\over 1 – 3 tan^2 A$ = tan 3A

$\implies$ L.H.S = $3\over tan 3A$ = 3 cot 3A = R.H.S