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

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

In a A B C ,\ A D is the bisector of the angle A meeting B C at Ddot If I is the incentre of the triangle, then A I\ : D I is equal to - (s in B+s in C):\ s in A (b) (cos B+cos C): cos A cos((B-C)/2):cos((B+C)/2) (d) sin((B-C)/2):sin((B+C)/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 : cosA +cosB+cosC+cosD=4 cos( (A+B)/2) cos((B+C)/(2) )cos( (C+A)/2)

cos(A-D)sin(B-C)+cos(B-D)sin(C-A)+cos(C-D)sin(A-B)=0

Prove in any triangle ABC that (i) a(cos B+cos C)=2(b+c) "sin"^(2)(A)/(2) (ii) a(cos C- cosB)= 2(b-c )"cos"^(2)(A)/(2)

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

If A+B-C=3pi,\ t h e n\ sin A+ sin B-sin C is equal to- a. 4sin(A/2)sin(B/2)cos(C/2) b. -4sin(A/2)sin(B/2)cos(C/2) c. \ 4cos(A/2)cos(B/2)cos(C/2) d. -4cos(A/2)cos(B/2)cos(C/2)