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Prove that : {20^{o}} . {40^{o}} . {60^{o}} . {80^{o}} = 3

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Prove that :
$tan 2 0^{o} . tan 4 0^{o} . tan 6 0^{o} . tan 8 0^{o} = 3$

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Prove that :

tan 2 0^{o} . tan 4 0^{o} . tan 6 0^{o} . tan 8 0^{o} = 3

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Text solutionVerified

tan 2 0^{\circ} tan 4 0^{\circ} tan 6 0^{\circ} tan 8 0^{\circ}

= tan 4 0^{\circ} tan 8 0^{\circ} tan 2 0^{\circ} tan 6 0^{\circ}

= \frac{sin 4 0 ^{\circ} sin 8 0 ^{\circ} sin 2 0 ^{\circ}}{cos 4 0 ^{\circ} cos 8 0 ^{\circ} cos 2 0 ^{\circ}} tan 6 0^{\circ}

= \frac{2 sin 4 0 ^{\circ} sin 8 0 ^{\circ} sin 2 0 ^{\circ}}{2 cos 4 0 ^{\circ} cos 8 0 ^{\circ} cos 2 0 ^{\circ}} (3)

= \frac{( cos 4 0 ^{\circ} - cos 12 0 ^{\circ} ) sin 2 0 ^{\circ}}{( cos 4 0 ^{\circ} + cos 12 0 ^{\circ} ) cos 2 0 ^{\circ}} (3)

= \frac{( 2 cos 4 0 ^{\circ} + 1 ) sin 2 0 ^{\circ}}{( 2 cos 4 0 ^{\circ} - 1 ) cos 2 0 ^{\circ}} (3)

= \frac{2 cos 4 0 ^{\circ} sin 2 0 ^{\circ} + sin 2 0 ^{\circ}}{2 cos 4 0 ^{\circ} cos 2 0 ^{\circ} - cos 2 0 ^{\circ}} (3)

= \frac{sin 6 0 ^{\circ} - sin 2 0 ^{\circ} + sin 2 0 ^{\circ}}{cos 6 0 ^{\circ} + cos 2 0 ^{\circ} - cos 2 0 ^{\circ}} 3

= tan 6 0^{\circ} \times 3

= 3 \times 3

= 3

Hence proved.

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Question Text	Prove that : $tan 2 0^{o} . tan 4 0^{o} . tan 6 0^{o} . tan 8 0^{o} = 3$
Answer Type	Text solution:1
Upvotes	150