The numerical value of tan 20- - tan 40- - -3tan 20-tan 40-2 must be -

tan 20^∘ + tan 40^∘ + √(3)tan 20^∘tan 40^∘ = sin 60^∘cos 20^∘cos 40^∘ + √(3)sin 20^∘sin 40^∘cos 20^∘cos 40^∘ = √(3)(12 + sin 20^∘sin 40^∘)cos 20^∘cos ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

The numerical...

Question

The numerical value of $(tan 20^{\circ} + tan 40^{\circ} + \sqrt{3} tan 20^{\circ} tan 40^{\circ})^{2}$ must be .....

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Solution

$tan 20^{\circ} + tan 40^{\circ} + \sqrt{3} tan 20^{\circ} tan 40^{\circ}$
$= \frac{sin 60^{\circ}}{cos 20^{\circ} cos 40^{\circ}} + \frac{\sqrt{3} sin 20^{\circ} sin 40^{\circ}}{cos 20^{\circ} cos 40^{\circ}} = \frac{\sqrt{3} (\frac{1}{2} + sin 20^{\circ} sin 40^{\circ})}{cos 20^{\circ} cos 40^{\circ}} = \sqrt{3}$

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$t a n 20^{\circ} + t a n 40^{\circ} + \sqrt{3} t a n 20^{\circ} t a n 40^{\circ}$ =

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The numerical value of (tan20∘+tan40∘+√3tan20∘tan40∘)2 must be .....

tan20∘+tan40∘+√3tan20∘tan40∘=sin60∘cos20∘cos40∘+√3sin20∘sin40∘cos20∘cos40∘=√3(12+sin20∘sin40∘)cos20∘cos40∘=√3

The numerical value of $(tan 20^{\circ} + tan 40^{\circ} + \sqrt{3} tan 20^{\circ} tan 40^{\circ})^{2}$ must be .....