Help proving these identities

1.) tanX-tanY / 1+tanX = cotY-cotX / cotXcotY+1

2.) sin5X + Sin3X / sin5X-sin3X = tan4X / tanX

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- Mar 10th 2010, 05:27 PMorochimaru700Proving Trig Identities
Help proving these identities

1.) tanX-tanY / 1+tanX = cotY-cotX / cotXcotY+1

2.) sin5X + Sin3X / sin5X-sin3X = tan4X / tanX - Mar 11th 2010, 12:15 AMGrandad
Hello orochimaru700Can I suggest first that:

I'm sure you mean for number (1):

- you should copy the question down carefully; and:
- either learn how to write these things using LaTeX (lots of information available on this web-site); or:
- use brackets correctly.

(tanX-tanY) / (1+tanXtanY) = (cotY-cotX) / (cotXcotY+1)Anyway, enough of my ranting. Divide top-and-bottom of the fraction on the LHS by $\displaystyle \tan X\tan Y$:

$\displaystyle \frac{\tan X - \tan Y}{1+ \tan X\tan Y}$$\displaystyle =\frac{\dfrac{\tan X}{\tan X\tan Y} - \dfrac{\tan Y}{\tan X\tan Y}}{\dfrac{1}{\tan X\tan Y}+ \dfrac{\tan X\tan Y}{\tan X\tan Y}}$

$\displaystyle =\frac{\dfrac{1}{\tan Y} - \dfrac{1}{\tan X}}{\dfrac{1}{\tan X}\dfrac{1}{\tan Y}+ \dfrac{1}{\tan X\tan Y}}$

$\displaystyle =\frac{\cot Y - \cot X}{\cot X \cot Y +1}$

For (2) use the sum-to-product formulae. For example, the numerator becomes $\displaystyle 2\sin 4x\cos x$. Use a similar formula on the denominator, simplify and you're there.

Can you complete this now?

Grandad - Mar 12th 2010, 04:57 PMorochimaru700
thnks for the advice and solution grandad, i'll try solve for # 2