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tanA,tanB,tanc are the roots of the eqn. x[p]3[/p]-k[p]2[/p]x[p]2[/p]-px+2k+1=0,then triangle abc will be a triangle of what nature?explain the ans?

tanA+tanB+tanC>0

since A+B+C=pi
=> tanA.tanB.tanC >0

SO 2 cases come
a) Tow factors negative and one factor positive
b) All three factors postive

but tan takes negative value when  pi>angle >pi/2  which is not possible (two angles cant be obtuse)
=> case a) rejected

=> all three factors positve

=> A,B,C <π/2

=> acute

[i]tan takes negative value when angle >pi [/i]

tan is negative when angle is in 2nd quad.

yes

ya...
@ eureka
tan is negative in 2nd & 4th quadrants.

hehe,,,ya,,,corrected now..

this is what happens when u reply after a looong time[3]

Shouldnt tan A + tan B + tan C = tan A tan B tan C for a triangle

But that would mean k[p]2[/p]+1 = 0 here!

edit: thought i was goofing there

k[p]2[/p] = -2k-1
=> (k+1)[p]2[/p] = 0
=> k = -1

so tanA + tanB + tanC = 1 = tanAtanBtanC

tanA +tanB+tanC ≥ 3 (iff tanA, tanB, tanC >0 )

but this is impossible

[hide]so one and only one of tanA, tanB and tanC is negative

so the triangle is obtuse[/hide]

edit: courtesy soumya:
tanA+tan+BtanC = 1. This implies that atleast one in tanA, tanB, tanC is negative but since tanAtanBtanC > 0 so two out of tanA, tanB, tanC is negative which is not possible for any triangle..

so no triangle possible

asish is right..

sorry i goofed up in another post.. din see the solution completely but pinked it after a couple of steps :(

whats the mistake in my soln??
I am very eager to know it...

since A+B+C=pi
=> tanA.tanB.tanC >0

how is this conclusion?

when A+B+C=pi
=>  tanA+tanB+tanC=tanA.tanB.tanC

proof
tan(A+B+C)=[frac]S1-S3[/]1-S2[/frac]
A+B+C=pi
=>tan(A+B+C)=0

=>S1=S3

=> tanA+tanB+tanC=tanA.tanB.tanC

Sir ,,,reached any conclusion ??

Ashis has written ---

tan a + tan b + tan c = 1 = tan a tan b tan c

tan a tan b tan c > 3 ( if tan a , tan b , tan c > 0 )

So one and only one of tan a , tan b , tan c is negative.

But that would imply that tan a tan b tan c < 0 , which is not correct

here as it is equal to 1 .

So either two of them are negative , which is not possible for a

triangle , or all of them are positive , which is again not possible as then

tan a tan b tan c > 1

So no such triangle is possible.

The reason that I am quite certain of my soln. , is that this ques. has

come up in one of Edudigm's quadratic practice papers , and the

answer given is no such triangle is possible .

can anyone tell me whats wrong in my soln ???

The wrong statement is ---

If a + b + c =pi , then tan a tan b tan c >0

take for example the angles 15 , 45 , and 120 . What do you get ? A negative number  !!!!

Also in an acute angled triangle , tan a tan b tan c > 3 , not > 0 , so you can't say that                

tan a + tan b + tan c = 1 , which you get if you solve k^2 = -2k -1

@soumya...
read post#12 and post#13 before any comments[1]

@akari  tanA +tanB +tanC >0

it wont be tanA +tanB +tanC≥0


btw nishnat sir plzz reply....the other day u  said my soln was rite...

That is what I am trying to say , in acute angled triangles , if tan a , tan b , tan c are all positive , then tan a tan b tan c > 3 . But here we get that tan a tan b tan c > 0 but < 3 . So no triangle is possible.

yes soumya.

ur correct

i should have written

tanA+tan+BtanC = 1. This implies that atleast one in tanA, tanB, tanC is negative but since tanAtanBtanC > 0 so two out of tanA, tanB, tanC is negative which is not possible for any triangle..