# Math Help - proving sum

1. ## proving sum

A+B+C = 180 (pi)
prove that,
cotA + cotB + cotC = cotA cotB cotC (1 + secAsecBsecC)

2. cotA + cotB + cotC = cotA cotB cotC (1 + secAsecBsecC)

= cotA cotB cotC + (cotA cotB cotC)(secAsecBsecC)

= cotA cotB cotC + (cos A/sin A)(cos B/sin B)(cos C/sin C)(1/(cosA cos B cos C))

cotA + cotB + cotC = (cot A cot B cot C) + (csc A csc B csc C)

= (cot A cot B cot (pi - A - B)) + (csc A csc B csc (pi - A - B))

= ?

3. Originally Posted by pacman
cotA + cotB + cotC = cotA cotB cotC (1 + secAsecBsecC)

= cotA cotB cotC + (cotA cotB cotC)(secAsecBsecC)

= cotA cotB cotC + (cos A/sin A)(cos B/sin B)(cos C/sin C)(1/(cosA cos B cos C))

cotA + cotB + cotC = (cot A cot B cot C) + (csc A csc B csc C)
after that,

cotA + cotB + cotC
= (cot A cot B cot C) + (csc A csc B csc C)
= (cosAcosBcosC / sinAsinBsinC) + (1/sinAsinBsinC)

taking 1/sinAsinBsinC outside the bracket,
= 1/sinAsinBsinC (cosAcosBcosC + 1) ....................(1)
____________________________________________

Now,
cotA + cotB + cotC
= cosA/sinA + cosB/sinB + cosC/sinC

cosAsinBsinC + cosBsinCsinA + cosCsinBsinA
= _________________________________________......... ........................(2)
sinAsinBsinC

Equating 1 and 2,

cosAsinBsinC + cosBsinCsinA + cosCsinBsinA
________________________________________ = 1/sinAsinBsinC (cosAcosBcosC +1)
sinAsinBsinC

sinAsinBsinC cancel out.

cosAsinBsinC + cosBsinCsinA + cosCsinBsinA = cosAcosBcosC +1

cosAsinBsinC + cosBsinCsinA + cosCsinBsinA - cosAcosBcosC = 1

(cosAsinBsinC + cosBsinCsinA) + (cosCsinBsinA - cosAcosBcosC) = 1

sinC(sinBcosA + cosBsinA) + cosC(sinBsinA - cosAcosB) = 1

sinC(sin(A+B)) + cosC(-(cosAcosB - sinBsinA)) = 1

sinC (sin (pi - C)) + cosC (- (cos(A+B)) =1

sinC (sinC) + cosC (-(-cos(pi-C)) =1

sinC^2 + cosC (-(-cosC)) =1

sinC^2 + cosC^2 =1

1=1

Hence proved

Thanks a bunch for kick-starting my brain. Just worked on it for a minutes and I got it.

4. Darknight, i believe it is wrong to prove an identity working on both sides because if you do, you are treating it as an equation not as an identity. Any piece of your thought on this?

5. ## Re:

Originally Posted by pacman
Darknight, i believe it is wrong to prove an identity working on both sides because if you do, you are treating it as an equation not as an identity. Any piece of your thought on this?
That's a good point, and you are correct: the proper way to
prove such identities is to begin on one side and algebraically
transform it into the form shown on the other side. Working on both
sides is technically incorrect because in doing so we are assuming
that the equality to be proven is already true.

However, in the sense that it is not mathematically incorrect - if
the identity is true, then the result of working on both sides will
eventually result in an equality that is "obviously" true.
That is, such a method won't lead to a wrong decision
on the truth or falseness of the identity in question. But in so far
as mathematical rigor, it is insufficient because of the reason
mentioned at the end of my first paragraph. Often, however, working on
both sides can help us form a rigorous proof of an identity. I will
illustrate this with the example.

Identity to be proven:

(1-TanA)/SecA + SecA/TanA = (1+TanA)/(SecATanA)

If we work on both sides, the first step is to multiply both sides by
Sec[a]Tan[a]:

(1-TanA)TanA + SecASecA = 1+TanA

Multiplying through, we find

TanA - TanA^2 + SecA^2 = 1+TanA,
or
-TanA^2 + SecA^2 = 1,

which is equivalent to the identity 1+TanA^2 = SecA^2, so we are
done. This is not a rigorous proof, but it leads to one, in which we
see the proper method is as follows:

(1-TanA)/SecA + SecA/TanA

= (1-TanA)TanA/ (SecATanA) + SecA^2/(SecATanA)

= (TanA - TanA^2 + SecA^2)/(SecATanA)

= (TanA - TanA^2 + 1 + TanA^2)/(SecATanA)

= (TanA + 1)/(SecATanA).

Therefore we have taken the left-hand side and transformed it into the
right-hand side using algebraic manipulation and other known
trigonometric identities. But it should not be too hard to see that
the basic "flow" of the proof is essentially the same procedure as the
"improper" method, just rearranged a bit.

I believe by retracing the steps it should be easy enough to prove my earlier identity, albeit it is much tougher than this simple example.

6. thanks darknight for that insightful explaination . . . .