`cos C - cos D = 2 "cos" (C - D)/(2) "cos"(C+D)/(2)`.

Assertion (A) : Suppose that alpha- beta is not an odd multiple of (pi)/(2) , m in R-{0,1) and (sin (alpha+beta))/(cos (alpha-beta))=(1-m)/(1+m) then Tan ((pi)/(4)-alpha)=m cot ((pi)/(4)- beta) Reason (R) : for all C,D, & R cos c+cos D= 2 cos ((C+D)/(2)) cos((C-D)/(2)) cos-cosD=-2sin ((C+d)/(2))sin ((C-D)/(2))

Prove that : cos C +cosD = 2 cos frac (C+D)(2) cos frac (C-D) (2) .

If A + B + C + D = 2pi , prove that: cos A + cos B + cos C + cos D =-4"cos"(A+B)/(2)"cos"(A+C)/(2)"cos"(A+D)/(2)

If A+B +C +D= 2 pi , prove that : cos A + cos B + cos C + cos D =-4 cos frac (A+B)(2) cos frac (A+C)(2) cos frac (A+D)(2) .

In a quadrilateral if (sin (A + B)) / (2) (cos (AB)) / (2) + (sin (c + D)) / (2) + (sin (c + D)) / ( 2) (cos (cD)) / (2) = 2, then (cos A) / (2) (cos B) / (2) + (cos A) / (2) (cos c) / (2) + (cos A) / (2) (cos D) / (2) + (cos B) / (2) (cos C) / (2) + (cos B) / (2) (cos D) / (2) + (cos C) / (2) (cos D) / (2)

If ABCD is a cyclic quadrilateral, then cos A + cos B is equal to............. A) 0 B) cos C + cos D C) - (cos C+ cos D) D) cos C- cos D

If (cos (AB)) / (cos (A + B)) + (cos (C + D)) / (cos (CD)) = 0 Prove that tan A tan B tan C tan D = -1

(cos (AB)) / (cos (A + B)) + (cos (C + D)) / (cos (CD)) = 0rArr tan A tan B tan C tan D =

Prove (i)sin C+sin D=2(sin(C+D))/(2)(cos(C-D))/(2)(ii)sin C-sin D=2(sin(C-D))/(2)(cos(C+D))/(2)

Prove that : sinC-sinD = 2 cos frac (C+D)(2) sin frac (C-D) (2) .