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Prove that for all triangles with angles $\alpha, \beta, \gamma$, $$\frac{\sin\alpha}{\cos\alpha + 1} + \frac{\sin\beta}{\cos\beta + 1} + \frac{\sin\gamma}{\cos\gamma + 1} = \frac{\cos\alpha + \cos\beta + \cos\gamma + 3}{\sin\alpha + \sin\beta + \sin\gamma}$$

Let $\tan\dfrac{\alpha}{2} = a, \tan\dfrac{\beta}{2} = b, \tan\dfrac{\gamma}{2} = c$, we have that $$\dfrac{\sin\beta}{\cos\beta + 1} = \dfrac{1}{b}, \cos\beta = \dfrac{1 - b^2}{1 + b^2}, \sin\beta = \dfrac{2b}{1 + b^2}$$ and $bc + ca + ab = 1$.

It needs to be proven that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{\dfrac{1 - a^2}{1 + a^2} + \dfrac{1 - b^2}{1 + b^2} + \dfrac{1 - c^2}{1 + c^2} + 3}{\dfrac{2a}{1 + a^2} + \dfrac{2b}{1 + b^2} + \dfrac{2c}{1 + c^2}}$$

$$\impliedby \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{\dfrac{1}{1 + a^2} + \dfrac{1}{1 + b^2} + \dfrac{1}{1 + c^2}}{\dfrac{a}{1 + a^2} + \dfrac{b}{1 + b^2} + \dfrac{c}{1 + c^2}}$$

$$\impliedby \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)\left(\frac{a}{1 + a^2} + \frac{b}{1 + b^2} + \frac{c}{1 + c^2}\right) = \frac{1}{1 + a^2} + \frac{1}{1 + b^2} + \frac{1}{1 + c^2}$$

$$\impliedby \left(\frac{a}{b} + \frac{a}{c}\right)\frac{1}{1 + a^2} + \left(\frac{b}{c} + \frac{b}{a}\right)\frac{1}{1 + b^2} + \left(\frac{c}{a} + \frac{c}{b}\right)\frac{1}{1 + c^2} = 0$$

$$\impliedby \frac{a(b + c)}{bc(c + a)(a + b)} + \frac{b(c + a)}{ca(a + b)(b + c)} + \frac{c(a + b)}{ab( b + c)(c + a)} = 0$$

$$\impliedby \frac{1 - bc}{(1 - ca)(1 - ab)} + \frac{1 - ca}{(1 - ab)(1 - bc)} + \frac{1 - ab}{(1 - bc)(1 - ca)} = 0$$

$$\impliedby (1 - bc)^2 + (1 - ca)^2 + (1 - ab)^2 = 0$$

$$\impliedby bc = ca = ab = 1 \impliedby bc + ca + ab = 3,$$ which is definitely incorrect.

I've surmised that the correct equality is $$\frac{\sin\alpha}{\cos\alpha + 1} + \frac{\sin\beta}{\cos\beta + 1} + \frac{\sin\gamma}{\cos\gamma + 1} = \frac{\cos\alpha + \cos\beta + \cos\gamma + 1}{\sin\alpha + \sin\beta + \sin\gamma},$$ but then I wouldn't know what to do first.

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    $\begingroup$ Isn't it $\frac{\sin{\alpha}}{1+\cos{\alpha}} = \frac{2\sin{\alpha/2}\cos{\alpha/2}}{1+2\cos^2{\alpha/2}-1} = \frac{\sin{\alpha/2}}{\cos{\alpha/2}} = \tan{\alpha/2}$ $\endgroup$ – HeatTheIce May 22 at 21:40
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Using your transformation I a get sligtly different equation: $$a+b+c=\frac{\sum_{cyc}\frac{1}{1+a^2}}{\sum_{cyc}\frac{a}{1+a^2}}$$ $$\sum_{cyc}\frac{a(a+b+c)}{1+a^2}=\sum_{cyc}\frac{1}{1+a^2}$$ $$\sum_{cyc}\frac{a^2+ab+ac-1}{(a+b)(a+c)}=0$$ $$\sum_{cyc}(a^2-bc)(b+c)=0$$ $$\sum_{cyc}a^2b+a^2c-b^2c-bc^2=0$$ Which is true!

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For $1+\cos\alpha\ne0,$ $$\dfrac{\sin\alpha}{1+\cos\alpha}=\cdots=\tan\dfrac\alpha2$$

Now, $$\tan\dfrac\alpha2+\tan\dfrac\beta2+\tan\dfrac\gamma2$$

$$=\dfrac{\sin\left(\dfrac\alpha2+\dfrac\beta2\right)}{\cos\dfrac\alpha2\cos\dfrac\beta2}+\dfrac{\sin\dfrac\gamma2}{\cos\dfrac\gamma2}$$

$$=\dfrac{\cos\dfrac\gamma2}{\cos\dfrac\alpha2\cos\dfrac\beta2}+\dfrac{\sin\dfrac\gamma2}{\cos\dfrac\gamma2}\text{ using }\alpha+\beta=\pi-\gamma$$

$$=\dfrac{\cos^2\dfrac\gamma2+\sin\dfrac\gamma2\cos\dfrac\alpha2\cos\dfrac\beta2}{\cos\dfrac\alpha2\cos\dfrac\beta2\cos\dfrac\gamma2}$$

Now the numerator

$$= 1-\sin^2\dfrac\gamma2+\sin\dfrac\gamma2\cos\dfrac\alpha2\cos\dfrac\beta2 $$

$$= 1-\sin\dfrac\gamma2\left(\sin\dfrac\gamma2-\cos\dfrac\alpha2\cos\dfrac\beta2\right) $$

$$= 1-\sin\dfrac\gamma2\left(\cos\left(\dfrac\alpha2+\dfrac\beta2\right) -\cos\dfrac\alpha2\cos\dfrac\beta2\right) $$

$$= 1+\sin\dfrac\alpha2\sin\dfrac\beta2\sin\dfrac\gamma2$$

Now use Prove trigonometry identity for $\cos A+\cos B+\cos C$

and If $A + B + C = \pi$, then show that $\sin(A) + \sin(B) + \sin(C) = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

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