Amath Trigo

[cot A cot B -1]/(cot A + cot B )
= [(1/tanA)(1/ tan B) -1]/(1/tan A + 1/tan B )
= [(1-tanAtanB)/tanA tan B) ]/((tanA+tanB)/(tan Atan B ))
= (1-tanAtanB)/(tanA+tanB)
=1/tan(A+B)
=cot (A+B)

2007-02-10 07:44:52 補充：
我要強調無人會寫若cot A + cot B=0則cot (A+B) = [cot A cot B -1]/(cot A + cot B) 不成立 因為大家心照﹐不需要寫出來的。你看A-MATH書都無寫這句

I suppose it is a standard formula that
　tan (A+B) = (tan A + tan B) / (1 - tan A tan B)
Then,
cot (A+B)
= 1 / tan (A+B)
= (1 - tan A tan B) / (tan B + tan A)
= (cot A cot B - 1) / (cot A + cot B),
　　by dividing the numerator and the denominator by cot A cot B.

2007-02-05 20:37:09 補充：
Typo... I mean, multiplying by cot A cot B, or dividing by tan A tan B -- for tan A tan B ≠ 0.Of course this answer has NOT YET finished, as we have to also consider the case oftan A tan B = 0, i.e. tan A = 0 or tan B = 0.

2007-02-05 20:43:04 補充：
However,when tan A tan B = 0,cot A cot B is undefined, and either cot A or cot B (or both) is undefined.In such case, the R.H.S. is undefined / undefined, so the equation becomes meaningless.So, the equation is set based on the condition that tan A tan B ≠ 0.

2007-02-05 20:43:18 補充：
--- answer finished ---