Question

sqrt(3) * tan theta * tan(theta + pi/3) * tan(theta + (2pi)/3) = 1​

Answer 1

Appropriate Question :-

Find general solution of

$\sf \: \sqrt{3} \: tan\theta \: tan(\bigg(\theta + \dfrac{\pi}{3}\bigg)tan\bigg(\theta + \dfrac{2\pi}{3}\bigg) = 1$

Identities Used :-

$\boxed{ \red{ \bf \:tan(x + y) = \dfrac{tanx + tany}{1 - tanxtany}}}$

$\boxed{ \red{ \bf \:tan\dfrac{\pi}{3} = \sqrt{3}}}$

$\boxed{ \red{ \bf \:tan\dfrac{2\pi}{3} = tan\bigg(\pi -\dfrac{\pi}{3}\bigg) = - tan\dfrac{\pi}{3} = -\sqrt{3}}}$

Let's solve the problem now!!!

Consider,

$\rm :\longmapsto\:tan\bigg(\theta + \dfrac{\pi}{3}\bigg)$

$\sf \: = \: \: \dfrac{tan\theta + tan\dfrac{\pi}{3} }{1 - tan\theta \: tan\dfrac{\pi}{3} }$

$\sf = \: \: \dfrac{tan\theta + \sqrt{3} }{1 - \sqrt{3}tan\theta}$

$\rm :\implies\:\boxed{ \red{ \bf \:tan\bigg(\theta + \dfrac{\pi}{3}\bigg) = \dfrac{tan\theta + \sqrt{3} }{1 - \sqrt{3}tan\theta}}}$

Consider,

$\rm :\longmapsto\:tan\bigg(\theta + \dfrac{2\pi}{3}\bigg)$

$\sf \: = \: \: \dfrac{tan\theta + tan\dfrac{2\pi}{3} }{1 - tan\theta \: tan\dfrac{2\pi}{3} }$

$\sf = \: \: \dfrac{tan\theta - \sqrt{3} }{1 + \sqrt{3}tan\theta}$

$\rm :\implies\:\boxed{ \red{ \bf \:tan\bigg(\theta + \dfrac{2\pi}{3}\bigg) = \dfrac{tan\theta - \sqrt{3} }{1 + \sqrt{3}tan\theta}}}$

Now,

Consider,

$\rm :\longmapsto\:\sf \: \sqrt{3} \: tan\theta \: tan(\bigg(\theta + \dfrac{\pi}{3}\bigg)tan\bigg(\theta + \dfrac{2\pi}{3}\bigg) = 1$

$\sf \: 1 = \: \sqrt{3} tan\theta \: \bigg( \dfrac{tan\theta + \sqrt{3} }{1 - \sqrt{3}tan\theta} \bigg)\bigg(\dfrac{tan\theta - \sqrt{3} }{1 + \sqrt{3}tan\theta} \bigg)$

$\sf \:1 = \: \sqrt{3}tan\theta\bigg(\dfrac{{tan}^{2}\theta - 3}{1 - {3tan}^{2}\theta }\bigg)$

$\sf \: 1= \: \sqrt{3}\bigg(\dfrac{{tan}^{3}\theta - 3tan\theta}{1 - {3tan}^{2}\theta }\bigg)$

$\sf \: 1 \: = \: - \sqrt{3}\bigg(\dfrac{3tan\theta - {tan}^{3}\theta}{1 - {3tan}^{2}\theta }\bigg)$

$\sf \: 1 \: = \: - \sqrt{3}tan3\theta$

$\rm :\implies\:tan3\theta = - \: \dfrac{1}{ \sqrt{3} }$

$\rm :\implies\:tan3\theta = tan\dfrac{5\pi}{6}$

$\rm :\implies\:3\theta = n\pi + \dfrac{5\pi}{6} \: where \: n \in \: Z$

$\bf :\implies\:\theta = \dfrac{n\pi}{3} + \dfrac{5\pi}{18} \: where \: n \in \: Z$

Additional Information:-

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)