if A+B+C = 180° , show that cot A cotB + cotBcotC +cotC cotA=1

if A+B+C = 180° ,
show that

tanA +tanB + tanC = tanAtanBtanC and

cot A cotB + cotBcotC +cotC cotA=1



given A+B+C = 180°

implies A + B = 180°- C

implies tan(A + B) = tan(180°- C)

using trigonometry formulae

tan(A + B) = -tan(C)

therefore

[tanA +tanB] / [1-tanAtanB ] = -tanC

cross multiplying

tanA +tanB = -tanC[1-tanAtanB ]

tanA +tanB = -tanC +tanAtanBtanC

or

tanA +tanB + tanC = tanAtanBtanC

divide each term with tanAtanBtanC we get


cot A cotB + cotBcotC +cotC cotA=1

some other problems
If A+B+C=π,
prove that: sinA+sinB+sinC = cot(A/2).cot(B/2) [sinA+sinB-sinC]

-----------------------------------------------------------
please leave your comments below
------------------------------------------------------------
index of math problems


disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work

some useful symbols

≡ ~
± ∓
∤ ◅
∈ ∉
⊆ ⊂ ∪ ∩ ⊥ ô
≈
≅
≠
≤
≥
·
∫ Σ → ∞
Π Δ Φ Ψ Ω Γ ∮ ∇∂ √ ∅ °
α β γ δ ε ζ η θ
ι κ λ μ ν ξ ο π
ρ σ τ υ φ χ ψ ω x²
some interesting symbols for posting into forums.


-----------------------------------------------------------
please leave your comments below
------------------------------------------------------------
index of math problems


disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work