To divide by zero in a trigonometric identity proof

Alright, so I did this test at school, and I was asked to prove this identity:

$\displaystyle \frac{tan(A) + tan(B)}{tan(A) - tan(B)} = \frac{sin(A+B)}{sin(A-B)}$

So I went along my merry way:

$\displaystyle LHS = \frac{\frac{sin(A)}{cos(A)} + \frac{sin(B)}{cos(B)}} {\frac{sin(A)}{cos(A)} - \frac{sin(B)}{cos(B)}}$

and prepared to make it into

$\displaystyle \frac{sin(A)cos(B) + cos(A)sin(B)}{sin(A)cos(B) - cos(A)sin(B)}$

But being the clueless person I am, I included the intermediary step

$\displaystyle = \frac{\frac{sin(A)cos(B)}{cos(A)}+sin(B)}{\frac{si n(A)cos(B)}{cos(A)} - sin(B)}$

which is wrong and loses marks even if it still reaches the answer.

The correct step in the given solution was to make a common denominator through:

$\displaystyle = \frac{\frac{sin(A)cos(B) + cos(A)sin(B)}{cos(A)cos(B)}}{\frac{sin(A)cos(B) - cos(A)sin(B)}{cos(A)cos(B)}}$

And the thing is, I don't understand what the difference between the two methods is. As far as I can see, the given correct solution multiplies both sides of the fraction by

$\displaystyle cos(A)cos(B)$

and I'm doing the same thing except multiplying by each cos(A) cos(B) one at a time instead of both at the same time. And as far as I can see, multiplying by two given numbers m and n is the same as multiplying by mn, right?

I asked my teacher about this and she said something about not accounting for cos(A) = 0. She justified the given correct solution with something like "the existence of tan(A) in the original solution automatically rules out cos(A) = 0 and that's why we can multiply both sides of it, but you made it into an addition and needed an extra condition 'for cos(A) = 0' in that line" (I don't remember exactly)

The proof concludes:

$\displaystyle LHS = \frac{sin(A)cos(B) + cos(A)sin(B)}{sin(A)cos(B) - cos(A)sin(B)}$

$\displaystyle = \frac{sin(A+B)}{sin(A-B)} (double angle formula)$

$\displaystyle = RHS $

$\displaystyle \therefore \frac{tan(A) + tan(B)}{tan(A) - tan(B)} = \frac{sin(A+B)}{sin(A-B)}$

Could someone shed some light on this? Thanks.

Re: To divide by zero in a trigonometric identity proof

Quote:

Originally Posted by

**icedtrees** Alright, so I did this test at school, and I was asked to prove this identity:

$\displaystyle \frac{tan(A) + tan(B)}{tan(A) - tan(B)} = \frac{sin(A+B)}{sin(A-B)}$

So I went along my merry way:

$\displaystyle LHS = \frac{\frac{sin(A)}{cos(A)} + \frac{sin(B)}{cos(B)}} {\frac{sin(A)}{cos(A)} - \frac{sin(B)}{cos(B)}}$

and prepared to make it into

$\displaystyle \frac{sin(A)cos(B) + cos(A)sin(B)}{sin(A)cos(B) - cos(A)sin(B)}$

But being the clueless person I am, I included the intermediary step

$\displaystyle = \frac{\frac{sin(A)cos(B)}{cos(A)}+sin(B)}{\frac{si n(A)cos(B)}{cos(A)} - sin(B)}$

which is wrong and loses marks even if it still reaches the answer.

The correct step in the given solution was to make a common denominator through:

$\displaystyle = \frac{\frac{sin(A)cos(B) + cos(A)sin(B)}{cos(A)cos(B)}}{\frac{sin(A)cos(B) - cos(A)sin(B)}{cos(A)cos(B)}}$

And the thing is, I don't understand what the difference between the two methods is. As far as I can see, the given correct solution multiplies both sides of the fraction by

$\displaystyle cos(A)cos(B)$

and I'm doing the same thing except multiplying by each cos(A) cos(B) one at a time instead of both at the same time. And as far as I can see, multiplying by two given numbers m and n is the same as multiplying by mn, right?

Yes that's correct.

I asked my teacher about this and she said something about not accounting for cos(A) = 0. She justified the given correct solution with something like "the existence of tan(A) in the original solution automatically rules out cos(A) = 0 and that's why we can multiply both sides of it, but you made it into an addition and needed an extra condition 'for cos(A) = 0' in that line" (I don't remember exactly)

The proof concludes:

$\displaystyle LHS = \frac{sin(A)cos(B) + cos(A)sin(B)}{sin(A)cos(B) - cos(A)sin(B)}$

$\displaystyle = \frac{sin(A+B)}{sin(A-B)} (double angle formula)$

$\displaystyle = RHS $

$\displaystyle \therefore \frac{tan(A) + tan(B)}{tan(A) - tan(B)} = \frac{sin(A+B)}{sin(A-B)}$

Could someone shed some light on this? Thanks.

Hi icedtrees,

Since $\displaystyle \tan A=\frac{\sin A}{\cos A}$ for $\displaystyle \tan A$ to have any meaningful value $\displaystyle \cos A\neq 0$. This is what your teacher had said. So in this case $\displaystyle \cos A\neq 0\mbox{ and }\cos B\neq 0$. Otherwise $\displaystyle \tan A \mbox{ and }\tan B$ would not exist. Is that clear to you?

Re: To divide by zero in a trigonometric identity proof

Thankyou Sudharaka, I understand what is meant by that now. But to clarify, proving it with

http://latex.codecogs.com/png.latex?...0sin%28B%29%7D

and then multiplying through by cos(A) should not be incorrect?

Re: To divide by zero in a trigonometric identity proof

Quote:

Originally Posted by

**icedtrees**

Whether you multiply by cos(A)cos(B) directly or by cos(A) and cos(B) in two steps there is no difference. Although the former is a little easier and reduces the number of steps.

Re: To divide by zero in a trigonometric identity proof

The identity has certain constraints on the domains of both sides.

On the right, we see that A and B must differ, due to the denominator.

More specifically, A-B must not be any multiple of 180 degrees,

otherwise the right side is undefined.

On the left, we see again that A and B must differ.

However, tanA and tanB are both undefined for any odd multiple of 90 degrees.

If you place A=90 degrees and B=30 degrees for example into the right side,

you get a solution, but try placing those angle values into the left!

Hence, tanA and tanB are both undefined when cosA and cosB are zero.

Therefore, multiplying by cosA/cosA=cosB/cosB=(cosAcosB)/(cosAcosB)=1

is not an issue,

since cosA and cosB cannot be zero for the allowable domain of this identity.

If you are aware of this, then your method is fine.