Monday

September 1, 2014

September 1, 2014

Posted by **Karen** on Monday, July 22, 2013 at 8:10pm.

- trig -
**Reiny**, Monday, July 22, 2013 at 8:34pmwe know that

tan 2A = 2tanA/(1 - tan^2 A)

so tan 210 = 2tan105/(1 - tan^2 105)

you should know that

tan 210 = tan 30° = 1/√3

let tan 105 = x

then

1/√3 = 2x/(1-x^2)

2√3x = 1 - x^2

x^2 + 2√3 - 1 = 0

x = (-2√3 ± √16)/2

= -√3 ± 2

but obviously tan105 is negative, ( 105° is in II )

so tan 105° = -√3-2

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