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

 

 

(tanx + tan45)/(1-(tanx)(tan45) = 3(3tanx – tan³x)/(1 - 3tan²x)              [identity of tan(a+b) and tan(3a)]

(tanx + 1)/(1-tanx) = (9tanx – 3tan³x)/(1 - 3tan²x)

now let tanx = a

(a+1)(1-3a²)=(9a-3a³)(1-a)

a+1-3a³-3a²=9a-3a³-9a²-3a⁴

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

 tanA + tanB + tanC + tanD = 0

hence proved

i hope this helps 

thanks