How to prove that -
$$\frac 1{\sin^2x} + \frac 1 {\sin^2y} - \frac{2\cos(x-y)}{\sin x\sin y} = \frac{\sin^2(x-y)}{\sin^2x\sin^2y}$$
How to prove that $1/\sin^2x + 1/\sin^2y - 2\cos(x-y)/(\sin x\sin y) = \sin^2(x-y)/(\sin^2x\sin^2y)$
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Does putting it all over a denominator of $\sin^2x\sin^2y$ and then simplifying the numerator not work? – Angina Seng Aug 19 '17 at 06:46
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@LordSharktheUnknown nope :( – Aug 19 '17 at 06:49
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You see, if that really doesn't work then the statement won't be true $\frown$. – Angina Seng Aug 19 '17 at 06:59
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i think this Statement is true, if the denominators are not equal zero – Dr. Sonnhard Graubner Aug 19 '17 at 07:03
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The numerator $$=\sin^2x+\sin^2y-\cos(x-y)\cdot2\sin x\sin y$$
$$=\sin^2x+\sin^2y-\cos(x-y)[\cos(x-y)-\cos(x+y)]$$
$$=\sin^2x+\sin^2y-\cos^2(x-y)+\cos(x-y)\cos(x+y)$$
Now use Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
lab bhattacharjee
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we have (the left-hand-side) $$\frac{\sin(x)^2+\sin(y)^2-2(\cos(x)\cos(y)+\sin(x)\sin(y))\sin(x)\sin(y)}{\sin(x)^2\sin(y)^2}$$ and this must be equal $$\frac{\sin(x)^2\cos(y)^2+\cos(x)^2\sin(y)^2-2\sin(x)\cos(x)\sin(y)\cos(y)}{\sin(x)^2\sin(y)^2}$$ now we will write only the numerators: both Terms are equal since $$2\sin(x)^2\sin(y)^2-2\left(\cos(x)\cos(y)+\sin(x)\sin(y)\right)\sin(x)\sin(y)=$$ $$-2\sin(x)\cos(x)\sin(y)\cos(y)$$
Dr. Sonnhard Graubner
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