Question

Prove that in easitest way,
$\tan(20) \tan(40) \tan(60) \tan(80) = 3$​

Answer 1

$\underline{ \underline{\large\green\maltese \large{ \bf \red{Given\: Equation:-}}}}$

$\sf\tan(20) \tan(40) \tan(60) \tan(80) = 3$

$\underline{ \underline{\large\purple\maltese \large{ \bf \orange{To\: Prove:-}}}}$

$\sf LHS=RHS$

$\underline{ \underline{\large\blue\maltese \large{ \bf \green{Solution:-}}}}$

$\underline{\underline{\large \sf Taking\:LHS}}$

$\sf\tan(20) \tan(40) \tan(60) \tan(80) =3$

$: \implies \tan(60) \tan(20) \tan(40) \tan(80) = 3$

$: \implies \sf\sqrt{3}. \tan(20) .\tan(60 - 20) \\.\tan(60 + 20)$

$: \implies$$\sqrt{3}. \tan(20) .(\frac{\tan60-\tan20}{1+\tan60.\tan20})$

$\quad\quad\quad.(\frac{\tan60+\tan20}{1-\tan60.\tan20})$

$: \implies$$\sf \sqrt{3}. \tan(20) .(\frac{ {tan}^{2} 60- {tan}^{2} 20}{1 - {tan}^{2} 60. {tan}^{2} 20})$

$: \implies$$\sf \sqrt{3}. \tan(20) .(\frac{ { (\sqrt{3}) }^{2} - {tan }^{2} 20}{1 - {(\sqrt{3}) }^{2}. {tan}^{2} 20})$

$: \implies$$\sf \sqrt{3}. \tan(20) .(\frac{ 3 - {tan }^{2}( 20)}{1 - {3.{tan}^{2} 20}})$

$: \implies$$\sf \sqrt{3}. (\frac{ 3 \tan(20) - {tan }^{3} (20)}{1 - {3.{tan}^{2} 20}})$

$: \implies$$\sf \sqrt{3} . \tan(3.(20))$

$: \implies$$\sf \sqrt{3} . \tan(60)$

$: \implies$$\sf \sqrt{3} . \sqrt{3}$

$: \implies$$\bf\red{3}$

$\sf\green{=LHS}$

$\underline{\underline{\mathfrak\pink{Hence \: Proved}}}$