Proving trignometrical identities: If $\sin(A+B) + \sin(B+C) + \cos(C-A) = \frac{3}{2}$ show that, $$1.\sin A + \cos B + \sin C = 0$$ $$2. \cos A + \sin B +\cos C = 0$$

The problem seems to be missing some assumptions, as noted by Americo.
For instance, If $B = 0$ and $A=C$ are acute angles, such that $\sin A  = 1/4$ the we have that
$\sin(A+B) + \sin(B+C) + \cos(A-C) = 3/2$, but none of
$\sin A + \cos B + \sin C$ or $\cos A + \sin B + \cos C$ are $0$.
In any case, this looks like a perfect problem for using complex numbers.
If $B' = \pi/2 - B$ and
$z_1 = \cos A + i \sin A$
$z_2 = \cos B' + i \sin B'$
$z_3 = \cos C + i \sin C$
The given identity is $\cos (A-B') + \cos (C - B') + \cos (A-C) = 3/2$
i.e.
$$\frac{z_1}{z_2} +\frac{z_2}{z_1} +\frac{z_3}{z_2} +\frac{z_2}{z_3} +\frac{z_1}{z_3} +\frac{z_3}{z_1}  = 3$$
The two identities


*

*$\sin A + \cos B + \sin C = 0$

*$\cos A + \sin B +\cos C = 0$
are equivalent to showing that $z_1 + z_2 + z_3 = 0$.
Eq 1, says that the imaginary part of $z_1 + z_2 + z_3$ is $0$ and Eq 2 says that the real part is $0$.
Hope that helps.
