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Trigonometry> prove that, tan20degrees+2tan40degrees+4c...

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prove that, tan20degrees+2tan40degrees+4cot80degrees=cot20degrees

Hasibul Islam , 7 Years ago

Grade 12th pass

2 Answers

Jayesh Gondhalekar

Last Activity: 7 Years ago

From table: tan20=0.3639 tan40= 0.839 cot80=0.1763. & cot20=2.7474 LHS: tan20+2 tan40+ 4 cot80= 0.3639+2(0.839)+4(0.1763)= =0.3639+1.678+0.7052=2.747=cot20 (Final answer)

Soumendu Majumdar

Last Activity: 7 Years ago

Dear Student,

L.H.S= $tan20^{\circ} + 2tan40^{\circ} + 4 cot80^{\circ}$

= $tan20^{\circ} + 2tan40^{\circ} + 4 [(cot^240^{\circ}-1)/2cot40^{\circ}]$ using cot 2A = $(cot^2 A -1)/2cotA$

= $tan20^{\circ} + (2 cot^2 40^{\circ} -2 +2tan40^{\circ}cot40^{\circ})/cot40^{\circ}$

= $tan20^{\circ} + 2 cot 40^{\circ}$

= $tan20^{\circ} + 2 [(cot^2 20^{\circ} - 1)/2cot20^{\circ}]$

= $[(tan20^{\circ}cot20^{\circ} +cot^2 20^{\circ} - 1)/cot20^{\circ}]$

= $cot^2 20^{\circ}/cot20^{\circ}$

= $cot20^{\circ}$ = R.H.S [Proved]

Hope it helps!

with regards,

Soumendu

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