1. ## Trigonometric Proof

Hi there,

I need help with proving two trigonometric expressions:

1) tanA - tanB = tanB tan(A-B)
tanA - cotB

2) tan (x - y) + tany
tan (x - y) - coty

2. Can anyone show me how to prove these questions?

3. hello friend, as far as i know, trig. expressions are not proven, they are usually simplified....are the above two questions identities that you want help with?

and with the second, if its proving, what do you want us to prove?

and for the first, do you want us to prove this:

$\frac{\tan A - \tan B} {\tan A - \cot B } = \tan B \tan (A-B)$

- sugarT

4. Originally Posted by Sallyvinka
Hi there,

I need help with proving two trigonometric expressions:

1) tanA - tanB = tanB tan(A-B)
tanA - cotB

2) tan (x - y) + tany
tan (x - y) - coty

In the first problem there is a typo. It should be

$\frac{tanA - tanB}{tanA + cotB}$

Put cot(B) = 1/tan(B) simplify.

In the second problem put cot(y) = 1/tan(y). The problem becomes

$\frac{tan(x-y) + tan(y)}{tan(x-y) - \frac{1}{tan(y)}}$

Now simplify.

5. The question in my textbook says prove that you can get from the left hand side to the right hand side using trigonometric equations for:

6. Originally Posted by Sallyvinka
The question in my textbook says prove that you can get from the left hand side to the right hand side using trigonometric equations for:

tan(A - B) = [( tan(A) - tan(B)]/[1 + tan(A)tan(B)]

So there must a print mistake in the problem.

7. ye there must, i tried working it out, it don't lead to what your textbook was asking

8. Hello Sallyvinka
Originally Posted by Sallyvinka
Hi there,

I need help with proving two trigonometric expressions:

1) tanA - tanB = tanB tan(A-B)
tanA - cotB
As others have pointed out, there is a mistake in this question. It should read:
$\frac{\tan A + \tan B}{\tan A - \cot B}=\tan B \tan(A-B)$
2) tan (x - y) + tany
tan (x - y) - coty
You don't say what you want to prove here. But perhaps it's this:
$\frac{\tan(x-y)+\tan y}{\tan(x-y) -\cot y}$
$=\frac{\tan y(\tan(x-y)+\tan y)}{\tan y\tan(x-y) -1}$, multiplying 'top-and-bottom' by $\tan y$

$=-\tan y\cdot\frac{\tan(x-y)+\tan y}{1-\tan y\tan(x-y)}$, re-arranging

$=-\tan y \cdot\tan([x-y]+y)$, using $\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$

$=-\tan x \tan y$