Here are all the results I got tan⁡(A+B+C)=a−b And (1+tan2A)(1+tan2B)(1+tan2C)=(1cos⁡Acos⁡Bcos⁡C)2 And cot⁡A+cot⁡B+cot⁡C=0 How should I...

pierdoodsu

pierdoodsu

Answered

2021-12-31

Here are all the results I got
tan(A+B+C)=ab
And
(1+tan2A)(1+tan2B)(1+tan2C)=(1cosAcosBcosC)2
And
cotA+cotB+cotC=0
How should I use these results?

Answer & Explanation

veiga34

veiga34

Expert

2022-01-01Added 32 answers

Write x=tanA,y=tanB,z=tanC
By Vieta’s formula, we have
x+y+z=a, xy+yz+xz=0,xyz=b
Now you want to find the value of
(1+x2)(1+y2)(1+z2)=1+x2+y2+z2+x2y2+y2z2+z2x2+x2y2z2
Note that
a2=(x+y+z)2=x2+y2+z2+2(xy+yz+zx)
so, x2+y2+z2=a2. Similarly,
0=(xy+yz+zx)2=x2y2+y2z2+z2x2+2xyz(x+y+z)
implies x2y2+y2z2+z2x2=2ab. Hence
(1+x2)(1+y2)(1+z2)=1+a2+2ab+b2=1+(a+b)2
lovagwb

lovagwb

Expert

2022-01-02Added 50 answers

Let p(x)=x3ax2+b, then
p(x)=(xtanA)(xtanB)(xtanC)
Now you need
p(i)p(i)
Vasquez

Vasquez

Expert

2022-01-08Added 457 answers

Let y=x2+1 so that x2=y1 and substitute as much as possible in the original cubic:
x(y1)a(y1)+b=0
Rearrange to isolate x and obtain:
x(y1)=a(y1)+b=ay(a+b)
Now square this:
x2(y1)2=(y1)3=(ay(a+b))2
so that
y33y2+3y+1a2y2+2a(a+b)y(a+b)2=0

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