SeftonIsAHole
Member
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- Jul 30, 2008
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- HSC
- 2009
Ok heres the question.
ABC is an acute-angled triangle.
i)Show that: tan(A+B)=-tanC
ii)Hence, show that: tanA+tanB+tanC=tanA.tanB.tanC
EDIT: Nvm, i got it lol
For (i) A+B+C = 180 (angle sum of a triangle)
therefore A+B=180-C
tan(A+B)=tan(180-C)
tan(A+B)=-tanC
(ii) tan(A+B)=-tanC
(tanA+tanB)/(1-tanAtanB)=-tanC
tanA+tanB+tanC=tanAtanBtanC
ABC is an acute-angled triangle.
i)Show that: tan(A+B)=-tanC
ii)Hence, show that: tanA+tanB+tanC=tanA.tanB.tanC
EDIT: Nvm, i got it lol
For (i) A+B+C = 180 (angle sum of a triangle)
therefore A+B=180-C
tan(A+B)=tan(180-C)
tan(A+B)=-tanC
(ii) tan(A+B)=-tanC
(tanA+tanB)/(1-tanAtanB)=-tanC
tanA+tanB+tanC=tanAtanBtanC