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

Thanks in advance!

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- May 17th 2010, 04:16 PMSallyvinkaTrigonometric 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

Thanks in advance! - May 17th 2010, 10:02 PMSallyvinka
Can anyone show me how to prove these questions?

- May 17th 2010, 10:35 PMsugarT
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:

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

- sugarT - May 17th 2010, 10:36 PMsa-ri-ga-ma
In the first problem there is a typo. It should be

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

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

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

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

Now simplify. - May 17th 2010, 10:44 PMSallyvinka
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, 10:51 PMsa-ri-ga-ma
- May 17th 2010, 10:55 PMsugarT
ye there must, i tried working it out, it don't lead to what your textbook was asking

- May 18th 2010, 08:08 AMGrandad
Hello SallyvinkaAs others have pointed out, there is a mistake in this question. It should read:

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

2)__tan (x - y) + tany__

tan (x - y) - coty

$\displaystyle \frac{\tan(x-y)+\tan y}{\tan(x-y) -\cot y}$Grandad$\displaystyle =\frac{\tan y(\tan(x-y)+\tan y)}{\tan y\tan(x-y) -1}$, multiplying 'top-and-bottom' by $\displaystyle \tan y$

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

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

$\displaystyle =-\tan x \tan y$

- May 21st 2010, 03:09 PMSallyvinka
I solved the question, thanks for everyone's help!