# Trigno formula error

• Jun 3rd 2008, 10:55 AM
omega2692
Trigno formula error
Guys ,i justed wanted to ask out something
if the formula given in NCERT is right
Tan(A+B)=Tan A+Tan B/1-Tan A*Tan B
then dont u think
{ Tan(A +B) }(1-Tan A*Tan B)= Tan A+Tan B
should be right too
but if put A & B =45
then
LHS=Not Defined(0)=Not Defined
&
RHS=1+1=2
whats wrong in formula then
• Jun 3rd 2008, 11:01 AM
Moo
Hello,

Quote:

Originally Posted by omega2692
Guys ,i justed wanted to ask out something
if the formula given in NCERT is right
Tan(A+B)=Tan A+Tan B/1-Tan A*Tan B
then dont u think
{ Tan(A +B) }(1-Tan A*Tan B)= Tan A+Tan B
should be right too
but if put A & B =45
then
LHS=Not Defined(0)=Not Defined
&
RHS=1+1=2
whats wrong in formula then

The formula is right :)

But when A=B=45, tanA=1 and tanB=1
Therefore, 1-tanA*tanB=0.

But you cannot multiply both sides this way by 0 :D
• Jun 3rd 2008, 11:01 AM
topsquark
Quote:

Originally Posted by omega2692
Guys ,i justed wanted to ask out something
if the formula given in NCERT is right
Tan(A+B)=Tan A+Tan B/1-Tan A*Tan B
then dont u think
{ Tan(A +B) }(1-Tan A*Tan B)= Tan A+Tan B
should be right too
but if put A & B =45
then
LHS=Not Defined(0)=Not Defined
&
RHS=1+1=2
whats wrong in formula then

The formula is right, but you need to be careful about the domain of the expressions.

For example:
$tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)~tan(B)}$
is clearly undefined for A = B = 45 as the denominator on the RHS is 0. Likewise the left hand side is undefined for A = B = 45, so no contradictions.

The result is that when you solve this for $tan(A) + tan(B)$:
$tan(A) + tan(B) = tan(A + B)~(1 - tan(A)~tan(B))$
and let A + B = 45, not only do you have an undefined quantity on the right hand side, but you have effectively multiplied both sides of the tan(A + B) equation by 0! So I wouldn't expect this equation to give any reasonable response for those values of A and B.

-Dan