Please help me prove this trig problem

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thechicinnovation

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tga x tgb + (tga + tgb) x ctg(a+b)=1
a- alpha
b- beta
x- times (multiplication)

I started out by opening up (tga + tgb) as tg(a+b), so I ended up with (tga x tgb)/(1-tga x tgb) and I turned ctg(a+b) into (ctga x ctgb -1)/(ctga + ctgb).
However, this really didn't help me, and now I'm stuck and don't know what I should do next.
 
thechicinnovation said:
tga x tgb + (tga + tgb) x ctg(a+b)=1
a- alpha
b- beta
x- times (multiplication)

I started out by opening up (tga + tgb) as tg(a+b), so I ended up with (tga x tgb)/(1-tga x tgb) and I turned ctg(a+b) into (ctga x ctgb -1)/(ctga + ctgb).
However, this really didn't help me, and now I'm stuck and don't know what I should do next.
Click to expand...

First, the way you wrote this is very confusing. Tan is used for tangent, not tg and cot for cotangent, not ctg. Also, do not use "x" when you mean multiplying. Either use parentheses or a dot. Using "x" for multiplication confuses it for the variable "x". That being said, you have:

\(\displaystyle [\tan(a)][\tan(b)]+[\tan(a)+\tan(b)][\cot(a+b)]=1\)

\(\displaystyle [\tan(a)+\tan(b)]\not=tan(a+b)\) as you state. In fact, you don't even have to monkey with any of the tan functions. But I would start by first expanding cot(a+b) as you said but you do it as 1/tan(a+b). If you do that correctly your solution should follow pretty quickly from there.
 
srmichael said:
First, the way you wrote this is very confusing. Tan is used for tangent, not tg and cot for cotangent, not ctg. Also, do not use "x" when you mean multiplying. Either use parentheses or a dot. Using "x" for multiplication confuses it for the variable "x". That being said, you have:

\(\displaystyle [\tan(a)][\tan(b)]+[\tan(a)+\tan(b)][\cot(a+b)]=1\)

\(\displaystyle [\tan(a)+\tan(b)]\not=tan(a+b)\) as you state. In fact, you don't even have to monkey with any of the tan functions. But I would start by first expanding cot(a+b) as you said but you do it as 1/tan(a+b). If you do that correctly your solution should follow pretty quickly from there.
Click to expand...

Sorry I write it like that, it's just that that's how we write it in Europe. But back to the point, how do you know that cot(a+b) equals 1/tan(a+b)?
 
thechicinnovation said:
Sorry I write it like that, it's just that that's how we write it in Europe. But back to the point, how do you know that cot(a+b) equals 1/tan(a+b)?
Click to expand...

Oh, I didn't know that's how the Euros worte that. Interesting.

Anyway, cot = 1/tan. That is one of the reciprocal identities that you should have learned. Just like csc = 1/sin and sec = 1/cos.
 
srmichael said:
Oh, I didn't know that's how the Euros worte that. Interesting.

Anyway, cot = 1/tan. That is one of the reciprocal identities that you should have learned. Just like csc = 1/sin and sec = 1/cos.
Click to expand...
okay, so that means that I can cross out both tan(a+b) and I will end up with (tana)(tanb) +1.. right?
If so, what should I do with the tana and tanb?
 
thechicinnovation said:
okay, so that means that I can cross out both tan(a+b) and I will end up with (tana)(tanb) +1.. right?
If so, what should I do with the tana and tanb?
Click to expand...

Close. You can cancel the [tan(a) + tan(b)] term and you are left with 1 - tan(a)tan(b). Combine this with the tan(a)tan(b) first term and wallah! Remember, \(\displaystyle \tan(a+b)=\dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\)
 
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