1.

If A+B+C+D = 360° then prove that tanA + tanB + tanC + tanD = 0PLEASE ANSWER IT AS FAST AS POSSIBLE. ..........

Answer»

I hope you know tan formula it goes like this -
tan(A+B) = (tanA+tanB)/1-tanAtanB

Now COMING to QUESTION,
Given A+B+C+D = 360°
use tan on both sides and separate using formula.

tan{(A+B)+(C+D)} = tan(360°)
=> tan(A+B) + tan(C+D) = 0.........eq (i)

=> tan(A+B) + (tanC+tanD)/1-tanCtanD = 0
=> tan(A+B) (1-tanCtanD) + tanC + tanD = 0

Using eq(i) in above result,

{tan(A+B) (1-tanCtanD) + tanC + tanD} = 0
=> -tan(C+D) (1-tanCtanD) + tanC + tanD = 0
=> - (tanC + tanD) + (tanC + tanD) = 0
=> 0 = 0
Hence Proved.

Please IGNORE if any typos. Though it is long, I hope it will help you.


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