Maths Question.. confused (1 Viewer)

[mrzeidan1]

ABC is an acute angled triangle

i) show that tan (a+b) = -tan c -----> done
ii) Hence show that tanA + tanB + tanC = tanA tanB tanC

I have a feeling its pretty easy, but I'm having major brainfreeze

thanks

 

[mrzeidan1]

Its OK I got it

 

[SoulSearcher]

I like those types of questions

 

[mrzeidan1]

Worth 3 marks in a past paper... Not a bad way to get a few marks

 

[xoxo]

can u explain it please? i did that same q....and also looked at it as though i had never seen it before...both parts....i feel so stupid lol

 

[Chocolat]

ABC is an acute angled triangle

i) show that tan (a+b) = -tan c -----> done
ii) Hence show that tanA + tanB + tanC = tanA tanB tanC

I have a feeling its pretty easy, but I'm having major brainfreeze

thanks

*stares blankly at the problem*

Oh. My. Goodness.

 

[mrzeidan1]

Ok I'll try before SoulSearcher runs to this thread

i) Angle sum of triangle, a+b+c =180
therefore a+b=180-c
tan (a+b) = tan (180-C) = -tan C

ii) tan(a+b) = (tan A + tan B) / (1- tan A tan B)
since tan(a+b) = -tan C,
(tan A + tan B) / (1- tan A tan B) = -tan C
multiplying by (1- tan A tan B), we get (tan A + tan B) = -tan C +tan A tan B tan C
therefore tan A + tan B + tan C = tan A tan B tan C

I really need to learn how to set this stuff out properly

 

[SoulSearcher]

I know another way to get out part 2, but it isn't nearly as nice and simple as that

 

[mrzeidan1]

share

 

[Chocolat]

Can someone please reassure me that this maths question is strictly in the 3 Unit Maths course (or 4 Unit) and not 2 Unit?

 

[mrzeidan1]

3 unit ...
I think/hope

 

[Riviet]

3 unit ...
I think/hope

Yeah it basically is. There's actually nothing that a 2 unit student wouldn't understand, it's just the algebraic manipulation that's harder.

 

[SoulSearcher]

Strictly 3 unit.

tan(A+B+C)
= tan((A+B)+C)
= tan(A+B)+tanC/1-tan(A+B)tanC
= [(tanA+tanB/1-tanAtanB)+tanC]/[1-(tanC(tanA+tabB)/1-tanAtanB)]
= [(tanA+tanB+tanC-tanAtanBtanC)/1-tanAtanB)]/[(1-tanAtanB-tanC(tanA+tanB))/1-tanAtanB]
= tanA+tanB+tanC-tanAtanBtanC/1-tanAtanB-tanC(tanA+tanB

However since A+B+C = 180, tan (A+B+C) = tan 180 = 0
Therefore tanA+tanB+tanC-tanAtanBtanC/1-tanAtanB-tanC(tanA+tanB) = 0
Therefore tanA+tanB+tanC-tanAtanBtanC = 0
Therefore tanA + tanB + tanC = tanA tanB tanC

See, a lot more messier than your way

 

[xoxo]

ohhhhhhhhh thanks lol it looks so easy now hahah

 

[mrzeidan1]

SoulSearcher made it more confusing lol

 

[SoulSearcher]

ohhhhhhhhh thanks lol it looks so easy now hahah

Told you my way was fucked

 

[fishy89sg]

please tell me where you got this

that combined question was worth 5 marks in our exam

 

[SoulSearcher]

I've seen it before in a couple of areas, can't remember where though.

 

[Riviet]

I've seen it before in a couple of areas, can't remember where though.

Was one of those places here?

 

[SoulSearcher]

Bingo!