Please help with this trig problem!!

Given triangle ABC,

sinC=(sinA+sinB)/(cosA+cosB),

What is the shape of triangle ABC? 

$\begin{array}{|rcll|} \hline \sin(C) &=& \dfrac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)} \\ && \boxed{\sin(A)+\sin(B) =2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} \\ && \boxed{\cos(A)+\cos(B) =2\cdot \cos\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} \\ \sin(C) &=& \dfrac{2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} {2\cdot \cos\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} \\ \sin(C) &=& \dfrac{ \sin\left(\dfrac{A+B}{2}\right) } { \cos\left(\dfrac{A+B}{2}\right) } \\ && \boxed{\sin(C)=\sin\Big(180^\circ-(A+B)\Big)=\sin(A+B)} \\ \sin(A+B) &=& \dfrac{ \sin\left(\dfrac{A+B}{2}\right) } { \cos\left(\dfrac{A+B}{2}\right) } \\ && \boxed{\sin(A+B) =2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A+B}{2}\right) } \\ 2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ \sin\left(\dfrac{A+B}{2}\right) } { \cos\left(\dfrac{A+B}{2}\right) } \\ 2\cdot \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ 1 } { \cos\left(\dfrac{A+B}{2}\right) } \\ \cos^2\left(\dfrac{A+B}{2}\right) &=& \dfrac{ 1 }{2} \\ \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ 1 }{\sqrt{2}} \\ \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ \sqrt{2} }{2} \\ \dfrac{A+B}{2} &=& \arccos\left( \dfrac{ \sqrt{2} }{2} \right) \\ \dfrac{A+B}{2} &=& 45^\circ \\ \mathbf{ A+B } & \mathbf{=} & \mathbf{ 90^\circ} \quad \text{ so } C=90^\circ \\ \hline \end{array}$

The shape is a right-angled triangle