$\tan A + \tan (60 ° + A) - \tan (60 ° - A) = 3 \tan 3 A$

Open in App

Solution

$LHS = \tan A + \tan (60 ° + A) - \tan (60 ° - A) = tanA + (\frac{\tan 60 ° + tanA}{1 - \tan 60 ° tanA}) - (\frac{\tan 60 ° - tanA}{1 + \tan 60 ° tanA}) [\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} and \tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}] = tanA + \frac{\sqrt{3} + tanA}{1 - \sqrt{3} tanA} - \frac{\sqrt{3} - tanA}{1 + \sqrt{3} tanA} = tanA + \frac{\sqrt{3} + 3 tanA + tanA + \sqrt{3} \tan^{2} A + \sqrt{3} + 3 tanA + tanA - \sqrt{3} \tan^{2} A}{(1 - \sqrt{3} tanA) (1 + \sqrt{3} tanA)} = tanA + \frac{8 tanA}{1 - 3 \tan^{2} A}$

$= \frac{tanA - 3 \tan^{3} A + 8 tanA}{1 - 3 \tan^{2} A} = \frac{9 tanA - 3 \tan^{3} A}{1 - 3 \tan^{2} A} = 3 (\frac{3 tanA - \tan^{3} A}{1 - 3 \tan^{2} A}) (∵ \tan 3 θ = \frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) = 3 \tan 3 A = RHS Hence proved .$

Suggest Corrections

Similar questions

tan A + tan (60^{\circ} + A) + tan (120^{\circ} + A) = 3 tan 3 A

Prove that following identities:

$t a n A + t a n (60^{\circ} + A) - t a n (60^{\circ} - A) = 3 t a n 3 A$

tanA+tan(60+A)+tan(60-A)=3tan3A

tan 40^{\circ} + tan 60^{\circ} + tan 80^{\circ} = tan 40^{\circ} . tan 60^{\circ} . tan 80^{\circ}

tan 60^{\circ} tan 80^{\circ} tan 20^{\circ} = tan 80^{\circ} - tan 20^{\circ} - tan 60^{\circ}

Join BYJU'S Learning Program