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I am trying to prove that,

$${\sin{A}\cos{A}-\sin{B}\cos{B}\over \sin^2{A}-\sin^2{B}}=\tan(90^o-A-B)$$

Using: $\sin(2A)=2\sin{A}\cos{A}$, then we have

$${1\over 2}{\sin(2A)-\sin(2B)\over \sin^2{A}-\sin^2{B}}=\tan(90^0-A-B)$$

$${1\over 2}{\sin(2A)-\sin(2B)\over (\sin{A}-\sin{B})(\sin{A}+\sin{B})}=\tan(90^0-A-B)$$

I am out of ideas, I need some help, anyone's?

2 Answers2

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$$ \frac{\sin A \cos A - \sin B \cos B}{\sin^2 A - \sin^2 B} = \frac{\sin (x-y) \cos (x+y)}{\sin (x-y) \sin(x+y)} = $$ $$ = \frac{\cos (x+y) }{\sin (x+y)} = \cot (x+y) = \tan (90 - x-y) $$

Olba12
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From where you set off:

$$ {1\over 2}{\sin(2A)-\sin(2B)\over \sin^2{A}-\sin^2{B}} \ = \ {1\over 2} \ \cdot \ {\frac{\sin(2A)-\sin(2B)}{[ \frac{1}{2} ( \ 1 - \cos[2A]) \ ] \ - \ [ \frac{1}{2} ( \ 1 - \cos[2B] \ ) \ ]}}$$

[applying "sine-squared identity"]

$$ = \ {1\over 2} \ \cdot \ {\frac{\sin(2A)-\sin(2B)}{ \frac{1}{2} ( \ \cos[2B] - \cos[2A] \ ) \ }} \ = \ {\frac{\sin(2A)-\sin(2B)}{ \ \cos(2B) - \cos(2A) }} $$

$$ = \ {\frac{2 \cos \left( \frac{2A+2B}{2} \right) \sin \left( \frac{2A-2B}{2} \right)}{ - 2 \sin \left( \frac{2B+2A}{2} \right) \sin \left( \frac{2B-2A}{2} \right) }} \ = \ {\frac{ \cos ( A + B) \sin \left( \frac{2A-2B}{2} \right)}{ \sin ( A + B) \sin \left( \frac{2A-2B}{2} \right) }} \ = \ {\frac{ \cos ( A + B) }{ \sin ( A + B) }} $$

[applying "sum-to-product" identities and odd symmetry property of sine function]

$$= \ \cot ( A + B) \ = \ \tan( \ 90º \ - \ [A+B] \ ) \ \ . $$

[This is not greatly different from Olba12's derivation, other than to proceed along the path you'd started on. There isn't really any getting around having to use either the product-to-sum or sum-to-product identities.]

EDIT (later that day) -- Here's a way to show this starting from the right-hand side:

$$ \tan( \ 90º \ - \ A \ - \ B \ ) \ = \ \cot ( A + B) \ = \ \frac{1}{ \tan ( A + B)} \ = \ \frac{1 \ - \ \tan A \ \tan B}{\tan A \ + \ \tan B} $$

$$ = \ \frac{1 \ - \ \tan A \ \tan B}{\tan A \ + \ \tan B} \ \cdot \ \frac{\tan A \ - \ \tan B}{\tan A \ - \ \tan B} \ = \ \frac{\tan A \ - \ \tan B \ - \ \tan^2 A \ \tan B \ + \ \tan A \ \tan^2 B}{\tan^2 A \ - \ \tan^2 B} $$

$$= \ \frac{(1 \ + \ \tan^2 B) \ \tan A \ - \ (1 \ + \ \tan^2 A) \ \tan B }{\tan^2 A \ - \ \tan^2 B} \ = \ \frac{\sec^2 B \ \tan A \ - \ \sec^2 A \ \tan B }{\tan^2 A \ - \ \tan^2 B} $$

$$ = \ \frac{\frac{\sin A}{\cos A \cos^2 B} \ - \ \frac{\sin B}{\cos B \cos^2 A} }{\frac{\sin^2 A}{\cos^2 A} \ - \ \frac{\sin^2 B}{\cos^2 B}} \ = \ \frac{ \frac{\sin A \cos A \ - \ \sin B \cos B}{\cos^2 A \cos^2 B} }{\frac{\sin^2 A \cos^2 B \ - \ \sin^2 B \cos^2 A}{\cos^2 A \cos^2 B}} $$

$$ = \ \frac{\sin A \cos A \ - \ \sin B \cos B}{\sin^2 A \cos^2 B \ - \ \sin^2 B \cos^2 A} \ = \ \frac{\sin A \cos A \ - \ \sin B \cos B}{\sin^2 A \ ( 1 - \sin^2 B) \ - \ \sin^2 B \ ( 1 - \sin^2 A)} $$

$$ = \ \frac{\sin A \cos A \ - \ \sin B \cos B}{\sin^2 A \ - \ \sin^2 A \sin^2 B \ - \ \sin^2 B \ + \ \sin^2 B \sin^2 A } \ = \ \frac{\sin A \cos A \ - \ \sin B \cos B}{\sin^2 A \ - \ \sin^2 B } \ \ . $$

All right, there is a way to do it without the more advanced identities -- it just requires rather more writing...

colormegone
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  • Well presented (+1) –  Jun 21 '16 at 01:21
  • Thank you. Your initial idea was a good one -- it becomes a matter of experience and familiarity with as many identities and relations as you can manage in order to "see" when you, say, want to factor an expression or manipulate it in a different way instead. (There are in fact a number of ways to establish the given equation.) Keep practicing! – colormegone Jun 21 '16 at 01:32
  • @RecklessReckoner, Suggesting the usage of http://math.stackexchange.com/questions/175143/prove-sinab-sina-b-sin2a-sin2b – lab bhattacharjee Jun 21 '16 at 05:34
  • Well, the thing about these "trig identity" problems is that there are a lot of different-looking routes one could take; had I done this during some other week, I might have made other choices. [I see that I end up arriving, in the denominator in my first approach, at a (slightly concealed) version of the identity in the post you link to.] Since OP was using the "double-angle formula" for sine, I decided to continue in that vein. – colormegone Jun 21 '16 at 05:54
  • One useful consequence of the angle-sum formulas, used in this A, is $\sin x -\sin y=2 \sin D \cos M,$ and $\cos x-\cos y= -2\sin D \sin M ,$ where$M=(x+y)/2$ and $D=(x-y)/2.$ – DanielWainfleet Jun 21 '16 at 06:20