1. ## Proving Trig Identities

Help proving these identities

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

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

2. Hello orochimaru700
Originally Posted by orochimaru700
Help proving these identities

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

2.) sin5X + Sin3X / sin5X-sin3X = tan4X / tanX
Can I suggest first that:

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

3. thnks for the advice and solution grandad, i'll try solve for # 2

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# (tan x tan y) (1- cot x cot y) (cot x cot y) (1- tan y tan y) = 0

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