The value f `tanA+tanB+tanC` is

Options

A. `45^(@),60^(@) and 75^(@)`
B. `30^(@),60^(@)and75^(@)`
C. `45^(@),30^(@)and75^(@)`
D. `75^(@),90^(@)and30^(@)`

Answer

A

Explanation

We have, `cosAcosBcosC=(sqrt(3)-1)/(8)`
and sin A sin B sin C = `(3+sqrt(3))/(8)`
`implies tanAtanBtanC=(3+sqrt(3))/(sqrt(3)-1)" ... (i)"`
But, `tanA+tanB+tanC=tanAtanBtanC`
`therefore tanA+tanB+tanC=(3+sqrt(3))/(sqrt(3)-1)" ... (ii)"`
Now, `A+B+C=pi`
`implies cos(A+B+C)=cospi`
`implies cosAcosBcosC-cosAsinBsinC-cosBsinCsinA-cosCsinAsinB=-1`
`implies cosAcosBcosC[1-tanBtanC-tanCtanA-tanAtanB]=1`
`implies(sqrt(3)-1)/(8){1-tanAtanB-tanBtanC-tanCtanA}=-1`
`impliestanAtanB+tanBtanC+tanCtanA=1+(8)/(sqrt(3)-1)`
`impliestanAtanB+tanBtanC+tanCtanA=5+4sqrt(3)" ... (iii)"`
From Eqs. (i), (ii), (iii), we find that tan A, tan B tan C are the roots of the equation.
`x^(3)-((3+sqrt(3))/(sqrt(3)-1))x^(2)+(5+4sqrt(3))x-((3+sqrt(3))/(sqrt(3)-1))=0`
or `x^(3)-(2+sqrt(3))sqrt(3)x^(2)+(5+4sqrt(3))x-(2+sqrt(3))sqrt(3)=0`
or `x^(3)-(3+2sqrt(3))x^(2)+(5+4sqrt(3))x-(3+2sqrt(3))=0`
Clearly, x = 1 is a root of the above equation. So, the given equation may be written as
`(x-1){x^(2)-(2+2sqrt(3))x+(3+2sqrt(3))}=0`
`implies(x-1)(x-sqrt(3)){x-(2+sqrt(3))}=0`
`impliesx=1,sqrt(3),2+sqrt(3)`
`impliestanA=1,tanB=sqrt(3)andtanC=2+sqrt(3)=(sqrt(3)+1)/(sqrt(3)-1)`
`impliesA=45^(@),B=60^(@) andC=75^(@)`
Hence, the angles of the triangle are of `45^(@), 60^(@) and 75^(@)`