SOLUTION: TANA+COTB/COTA+TANB=TANA+COTB

Algebra.Com
Question 470938: TANA+COTB/COTA+TANB=TANA+COTB
Answer by Edwin McCravy(20064)   (Show Source): You can put this solution on YOUR website!
TANA+COTB/COTA+TANB=TANA+COTB
You didn't put in any parentheses to show where the
numerators and denominators begin and end.

However this equation cannot be an identity regardless
of where the parentheses should go -- because
suppose A=45° and B=45°.  Then 

tan(A) = tan(B) = cot(A) = cot(B) = 1 

The above becomes:

TANA+COTB/COTA+TANB=TANA+COTB

           1+1/1+1 = 1+1

Theis is no way to put parentheses in the left side to
make that equation true.

Edwin

RELATED QUESTIONS

cotA+cotB/tanA+tanB=cotB/tanA (answered by Tatiana_Stebko)
tanA+cotB/tanB+cotA=tanA*cotB (answered by Edwin McCravy)
CotA+cotB/tanA+cotB=cot/tanB (answered by Alan3354)
CotA+tanB/tanA+cotB=cotA/tanB (answered by Alan3354)
show that cotA + tanB / cotB + tanA... (answered by hamsanash1981@gmail.com)
Prove that (tanA - tanB)/(1 - tanAtanB) = (cotB - cotA)/cotBcotA - 1 (answered by MathLover1)
Which one of the following statements is true for trigonometric functions of two acute... (answered by vleith)
Prove that:~ cotB + tanB = cotB... (answered by Alan3354)
(tanA)(cotA) (answered by jim_thompson5910)