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 use these results?

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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 use these results?

veiga34 · Accepted Answer

Write x=tan⁡A,y=tan⁡B,z=tan⁡C   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 · Accepted Answer

Let p(x)=x3−ax2+b, then   p(x)=(x−tan⁡A)(x−tan⁡B)(x−tan⁡C)   Now you need  p(i)⋅p(−i)

Vasquez · Accepted Answer

Let y=x2+1 so that x2=y−1 and substitute as much as possible in the original cubic: x(y−1)−a(y−1)+b=0 Rearrange to isolate x and obtain: x(y−1)=a(y−1)+b=ay−(a+b) Now square this: x2(y−1)2=(y−1)3=(ay−(a+b))2 so that y3−3y2+3y+1−a2y2+2a(a+b)y−(a+b)2=0

Here are all the results I got \tan(A+B+C)=a-b And (1+\tan^2 A)(1+\tan^2 B)(1+\tan^2 C)=(\frac{1}{\cos

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