Question

If $A,B,C$ are the angles of a triangle, show that
$\frac{\cos A\cos C+\cos (A+B)\cos (B+C)}{\cos A\sin C-\sin (A+B)\cos (B+C)}=\cot C$

Answer 1

As we know that the sum of all thr angles of a triangle is 180 degree.

$A+B+C=180A+B=180-CB+C=180-C$

$\frac{\cos A\cos C+\cos (A+B)\cos (B+C)}{\cos A\sin C-\sin (A+B)\cos (B+C)}=\frac{\cos A\cos C+\cos (180-C)\cos (180-A)}{\cos A\sin C-\sin (180-C)\cos (180-A)}=\frac{\cos A\cos C+\cos \left(A\right)\cos \left(C\right)}{\cos A\sin C+\sin \left(C\right)\cos \left(A\right)}=\frac{2\cos A\cos C}{2\cos A\sin C}=\cot C$