Trigonometric Proof

• May 17th 2010, 05:16 PM
Sallyvinka
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

• May 17th 2010, 11:02 PM
Sallyvinka
Can anyone show me how to prove these questions?
• May 17th 2010, 11:35 PM
sugarT
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
• May 17th 2010, 11:36 PM
sa-ri-ga-ma
Quote:

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.
• May 17th 2010, 11:44 PM
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:

http://www.mathhelpforum.com/math-he...b2104a65-1.gif
• May 17th 2010, 11:51 PM
sa-ri-ga-ma
Quote:

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:

http://www.mathhelpforum.com/math-he...b2104a65-1.gif

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

So there must a print mistake in the problem.
• May 17th 2010, 11:55 PM
sugarT
ye there must, i tried working it out, it don't lead to what your textbook was asking
• May 18th 2010, 09:08 AM
Hello Sallyvinka
Quote:

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

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 $\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$

$=-\tan x \tan y$