In any quadrilateral ABCD, show that: `cosA cosB-cosC cosD="sin"A "sin"B-"sin"C "sin"D.`

In a cyclic quadrilateral ABCD, then find cosA+cosB+cosC+cosD=?

In a quadrilateral ABCD,cos A cos B+sin C sin D=

If A, B, C, D be the angles of acyclic quadrilateral, show that : cosA +cosB+cosC+cosD=0 .

In a Quadrilateral ABCD,cos A*cos B+sin C sin D=

If any quadrilateral ABCD,prove that sin(A+B)+sin(C+D)=0cos(A+B)=cos(C+D)

Show that: sin A sin(B-C)+sin B sin(C-A)+sin C sin(A-B)=0

If A,B,C and D are the angles of cyclic quadrilateral, prove that: i) cosA + cosB+cosC+cosD ii) cos(180^(@)-A)+cos(180^(@)+B)+cos(180^(@)+C)-sin(90^(@))

If A+B+C=pi , prove that : cosA + cosB-cosC=4cos(A/2) cos(B/2) sin(C/2) -1

show that sin A*sin(B-C)+sin B*sin(C-A)+sin C*sin(A-B)=0