三角形的三个角记为A, B, C
证明tan(A)tan(B)tan(C) = tan(A) + tan(B) + tan(C)
这个初中平面几何怎么做？
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#2 Re: 这个初中平面几何怎么做？
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The statement tan(A)tan(B)tan(C) = tan(A) + tan(B) + tan(C) is true when A, B, and C are the interior angles of a triangle. This is because A + B + C = π, and tan(π - x) = -tan(x).
Proof:
1. Start with the given equation:
.
A + B + C = π.
2. Express tan(C) in terms of A and B:
.
Since A + B + C = π, then C = π - (A + B). Therefore, tan(C) = tan(π - (A + B)).
3. Use the tangent of a sum formula:
.
tan(π - (A + B)) = -tan(A + B) = - (tan(A) + tan(B)) / (1 - tan(A)tan(B)).
4. Substitute tan(C) back into the original equation:
.
tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C) becomes tan(A) + tan(B) - (tan(A) + tan(B)) / (1 - tan(A)tan(B)) = tan(A)tan(B) * - (tan(A) + tan(B)) / (1 - tan(A)tan(B)).
5. Simplify the equation:
.
After some algebraic manipulation, you will arrive at tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C).
This video demonstrates how to prove that tanA + tanB + tanC = tanAtanBtanC:
Learn more
The statement tan(A)tan(B)tan(C) = tan(A) + tan(B) + tan(C) is true when A, B, and C are the interior angles of a triangle. This is because A + B + C = π, and tan(π - x) = -tan(x).
Proof:
1. Start with the given equation:
.
A + B + C = π.
2. Express tan(C) in terms of A and B:
.
Since A + B + C = π, then C = π - (A + B). Therefore, tan(C) = tan(π - (A + B)).
3. Use the tangent of a sum formula:
.
tan(π - (A + B)) = -tan(A + B) = - (tan(A) + tan(B)) / (1 - tan(A)tan(B)).
4. Substitute tan(C) back into the original equation:
.
tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C) becomes tan(A) + tan(B) - (tan(A) + tan(B)) / (1 - tan(A)tan(B)) = tan(A)tan(B) * - (tan(A) + tan(B)) / (1 - tan(A)tan(B)).
5. Simplify the equation:
.
After some algebraic manipulation, you will arrive at tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C).
This video demonstrates how to prove that tanA + tanB + tanC = tanAtanBtanC:
#5 Re: 这个初中平面几何怎么做？
初中解法呢？画辅助线那种agi 写了： 2025年 5月 15日 21:29 [AI Overview
Learn more
The statement tan(A)tan(B)tan(C) = tan(A) + tan(B) + tan(C) is true when A, B, and C are the interior angles of a triangle. This is because A + B + C = π, and tan(π - x) = -tan(x).
Proof:
1. Start with the given equation:
.
A + B + C = π.
2. Express tan(C) in terms of A and B:
.
Since A + B + C = π, then C = π - (A + B). Therefore, tan(C) = tan(π - (A + B)).
3. Use the tangent of a sum formula:
.
tan(π - (A + B)) = -tan(A + B) = - (tan(A) + tan(B)) / (1 - tan(A)tan(B)).
4. Substitute tan(C) back into the original equation:
.
tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C) becomes tan(A) + tan(B) - (tan(A) + tan(B)) / (1 - tan(A)tan(B)) = tan(A)tan(B) * - (tan(A) + tan(B)) / (1 - tan(A)tan(B)).
5. Simplify the equation:
.
After some algebraic manipulation, you will arrive at tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C).
This video demonstrates how to prove that tanA + tanB + tanC = tanAtanBtanC:
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