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If tanα.tanβ = a and α + β = π/6 then tanα and tanβ are the roots of the...

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nasirjam | Student, Grade 9 | Honors

Posted April 8, 2013 at 11:18 AM via web

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If tanα.tanβ = a and α + β = π/6 then tanα and tanβ are the roots of the quadratic equation:

a. sqrt(3)x^2 - (1 - a)x +sqrt(3)a = 0

b. sqrt(3)x^2 - (1 + a)x + sqrt(3)a = 0

c. sqrt(3)x^2 + (1 - a)x - sqrt(3)a = 0

d. sqrt(3)x^2 + (1 - a)x + sqrt(3)a = 0

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pramodpandey | College Teacher | Valedictorian

Posted April 8, 2013 at 11:42 AM (Answer #1)

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`tan(alpha+beta)=(tan (alpha)+tan(beta))/(1-tan(alpha)tan(beta))`

`tan(pi/6)=(x+y)/(1-a)`

where  x=`tan(alpha) ,y=tan(beta)`

`x+y=(1-a)/sqrt(3)`

`xy=a`

Ans. a.

sum of the roots=(1-a)/sqrt(3)

product of the roots= a

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oldnick | Valedictorian

Posted April 8, 2013 at 11:29 PM (Answer #2)

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a) First divide the equations by sqrt(3)

x^2 - (1-a)x /sqrt(3) - a = 0


so:   tg`alpha`  + tg `beta`  = 1-  tg `beta`tg`alpha` /sqrt(3)

thus  (tg`alpha` +tg`beta` )/( 1 -tg`alpha` tg`beta` ) = sqrt(3)/3

that is:  tg( `alpha`+ `beta` )= sqrt(3)/3

and verifies for + `` = `pi` /6

on the other side a = tg`alpha` tg`beta`
so are the solutions.

b)  x^2 - (1 + a) x /sqrt(3)+ a = 0

   At the same:  tg `alpha` +tg `beta` = (1 +tg`alpha` tg`beta` )/sqrt(3)

so [that tg `alpha` -(-tg`beta` )]/ [1 -tg`alpha` (-tg`beta` )]  =sqrt(3)/3

       tg(`alpha` + `pi` -`beta` )=sqrt(3)  /3

 tg(5`pi` /6) = sqrt(3)/3 not verified

c) tg`alpha` +tg`beta` =-(1 - tg`alpha` tg`beta` )/sqrt( 3)

then:

-(tg `alpha` + tg `beta` ) =-( 1 - tg`alpha` tg`beta` )/sqrt(3)

tg(`alpha` +`beta` )= sqrt(3)/3 so is veriifed


d) - (tg`alpha` +tg`beta` )= 1 - tg`alpha`  tg`beta` /sqrt(3)

so (as the point "c", :

    - tg(`alpha` +`beta` )= sqrt(3)/3

       tg[`pi` -(`alpha` +`beta` )]=tg(5`pi` /6)= sqrt(3)/3

not verified

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