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# if A,B,C,D are solutions of tan(x+45deg)=3tan3x then show that tanA+tanB+tanC+tanD=0

Ankit Jaiswal
165 Points
5 years ago
the values can be found easily by the help of trigo identities and some properties of polynomials

tan(x+45)=3tan(3x)

$\Rightarrow$(tanx + tan45$\degree$)/(1-(tanx)(tan45$\degree$) = 3(3tanx – tan3x)/(1 - 3tan2x)              [identity of tan(a+b) and tan(3a)]
$\Rightarrow$(tanx + 1)/(1-tanx) = (9tanx – 3tan3x)/(1 - 3tan2x)
now let tanx = a
$\therefore$ (a+1)/(1-a)=(9a-3a3)/(1-3a2)
$\Rightarrow$(a+1)(1-3a2)=(9a-3a3)(1-a)
$\Rightarrow$a+1-3a3-3a2=9a-3a3-9a2-3a4
$\Rightarrow$3a4-0a3+6a2-8a+1=0      [ the solutions or roots of this equation will be tanA,tanB,tanC,tanD as the roots of the question is given as x = A or B or C or D and a=tanx here]
now we can compare this equation as
px4+qx3+rx2+sx+t
and we know that the sum of solutions or roots of this equaion is
-q/p and here -q=0
hence sum of roots equal 0
$\therefore$ tanA + tanB + tanC + tanD = 0
hence proved
i hope this helps
thanks