We have,<\/p>\n
L.H.S = cot A + cot (60 + A) – cot (60 – A)<\/p>\n
\\(\\implies\\) L.H.S = \\(1\\over tan A\\) + \\(1\\over tan (60 + A)\\) – \\(1\\over tan (60 – A)\\)<\/p>\n
\\(\\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\\)<\/p>\n
[ By using this formula, tan (A + B) = \\(tan A + tan B\\over 1 – tan A tan B\\) above ]<\/p>\n
\\(\\implies\\) L.H.S = \\(1\\over tan A\\) – \\(8 tan A\\over 3 – tan^2 A\\)<\/p>\n
\\(\\implies\\) L.H.S = \\(3 – 9 tan^2 A\\over 3 tan A – tan^3 A\\)<\/p>\n
L.H.S = 3(\\(1 – 3 tan^2 A\\over 3 tan A – tan^3 A\\))<\/p>\n
Since \\(3 tan A – tan^3 A\\over 1 – 3 tan^2 A\\) = tan 3A<\/p>\n
\\(\\implies\\) L.H.S = \\(3\\over tan 3A\\) = 3 cot 3A = R.H.S<\/p>\n","protected":false},"excerpt":{"rendered":"
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} …<\/p>\n