How Does Tan A + Tan C Equal b^2/ac in a Right Triangle?

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Discussion Overview

The discussion revolves around the mathematical relationship involving the tangent of angles in a right triangle, specifically the equation tan A + tan C = b² / ac. Participants explore the derivation of this equation, the application of the Pythagorean theorem, and the relationships between the sides of the triangle.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants express that tan A = a/c and tan C = c/a, leading to the expression (c²)(a²)/ac when attempting to add the tangents.
  • There is confusion regarding how addition of tangents translates into multiplication in the derived expression.
  • Some participants reference the Pythagorean theorem, suggesting that for the equation to hold, a² must be subtracted from c² to yield b².
  • One participant clarifies that the Pythagorean theorem states a² + b² = c², indicating that b is the hypotenuse.
  • Another participant challenges the naming convention of the triangle's sides, noting that the designation of sides can affect the interpretation of the theorem.
  • There is a suggestion that b² could equal c² - a², indicating a potential misunderstanding of the triangle's configuration.

Areas of Agreement / Disagreement

Participants do not reach consensus on the relationships between the sides of the triangle or the validity of the derived equation. Confusion remains regarding the application of the Pythagorean theorem and the definitions of the triangle's sides.

Contextual Notes

Participants express uncertainty about the implications of the Pythagorean theorem in this context and how the sides are defined in relation to the angles A, B, and C.

wellY--3
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ABC is right angled at B.

Prove that tanA+tanC=b^2 / ac


Ok so far I've got that tan A= a/c and tanC=c/a and then when i try to add those together i get (c^2)(a^2)/ac , but how can (c^2)(a^2) = b^2
all I've thought of is pythagorus's theorem. But for it to equal b^2, a^2 has to be taken away from c^2
 
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wellY--3 said:
ABC is right angled at B.

Prove that tanA+tanC=b^2 / ac


Ok so far I've got that tan A= a/c and tanC=c/a and then when i try to add those together i get (c^2)(a^2)/ac

Are you sure about this? How did adding become multiplication?

wellY--3 said:
but how can (c^2)(a^2) = b^2

It doesn't

wellY--3 said:
all I've thought of is pythagorus's theorem. But for it to equal b^2, a^2 has to be taken away from c^2

Again are you sure? The Pythagorean theorem says that a2+b2=c2 where c is the hypotenuse of the right triangle and a and b are the two legs of the triangle. Is that the case in this problem?
 
wellY--3 said:
ABC is right angled at B.

Prove that tanA+tanC=b^2 / ac


Ok so far I've got that tan A= a/c and tanC=c/a and then when i try to add those together i get (c^2)(a^2)/ac , but how can (c^2)(a^2) = b^2
all I've thought of is pythagorus's theorem. But for it to equal b^2, a^2 has to be taken away from c^2

Note: [tex]\frac{a}{c}+\frac{c}{a} = \frac{a^2+c^2}{a c}[/tex]
 
ok ok so then it equals c^2+a^2/ca
I still don't see how the top can equal b^2
 
did someone mentioned that this triangle is right-angled?
 
wellY--3 said:
ok ok so then it equals c^2+a^2/ca
I still don't see how the top can equal b^2
by Pythagorean theorem, a2 + c2 = b2 for the triangle you mentioned. B is the right angle here. so b is the hypotenuse. so it's square should be equal to the sum of the squares of the other two sides, namely a and c
 
i thought c was always the hypotenuse ...so b^2=c^2-a^2
 
wellY--3 said:
i thought c was always the hypotenuse ...so b^2=c^2-a^2

It depends what you call the sides! In your first post, you appear to be adopting the naming that calls the sides opposite angles A, B and C, a, b and c respectively.

Note that Pythagoras' Theorem simply says that the square of the hypotenuse is equal to the sum of the squares of the other two sides. Thus, in this case, b2=a2+c2
 

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