Prove that for all $\alpha + \beta + \gamma = \pi$, $\sum_{cyc}\frac{\sin\beta}{\cos\beta + 1} = \frac{\sum_{cyc}\cos\beta + 3}{\sum_{cyc}\sin\beta}$.

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 $\sin(A) + \sin(B) + \sin(C) = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$