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Given a line $r$ and a (superior) semicircle perpendicular to $r$, and an arc $[AB]$ in the semicircle, where $\alpha$ is the angle $A'ÔA$ and $\beta$ is the angle $B'ÔB$.

I need to proof the following:

$ \frac{ \frac{AA'. BB'}{AB'. BA'} - \frac{1}{\frac{AA'. BB'}{AB'. BA'}}}{2} = \frac{\cos{\alpha}+\cos{\beta}}{\sin{\alpha}\sin{\beta}} $

I can't do it because I can't write the right side of the equation without using measures of segments that don't appear in the other side. I tried cosine law, sine law, and other manipulations, but I'm starting to believe I need to go further than using trigonometric idenitities. Can someone give me a hint? Thanks.

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    $\begingroup$ This proof is similar, and using it the problem will be only writing the left side of the current problem as the expression on the third side of the first equation line of link problem. math.stackexchange.com/questions/1689160/… $\endgroup$
    – user286485
    Mar 9 '16 at 0:34
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This proof is similar, and using it the problem will be only writing the left side of the current problem as the expression on the third side of the first equation line of link problem.

This can be done simply merging the two fractions in the numerator and isolating $2$.

Proof of an identity that relates hyperbolic trigonometric function to an expression with euclidean trigonometric functions.

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