Suppose that `alpha-beta` is not an odd multiple of `(pi)/2, m in R-{0,-1}` an `(sin (alpha+beta))/(cos(alpha-beta))=(1-m)/(1+m)`. Then show that `tan((pi)/4-alpha)=mtan((pi)/4+beta)`

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))

If (sin (theta+ alpha))/(cos (theta- alpha))=(1-M)/(1+M) then Tan ((pi)/(4)- theta). Tan ((pi)/(4)- alpha)=

If ( sin ( theta + alpha ))/( cos ( theta - alpha ))=(1- m)/(1+m) , " then " tan"((pi)/4- theta )=

If sin (alpha+beta ) =1 , sin (alpha-beta ) =1//2 " then " tan (alpha +2beta ) tan(2alpha+beta ) =

If cos^(-1)[(cos alpha+ cos beta)/(1+cos alpha cos beta)]=2 Tan^(-1)x then x=

If 3 sin alpha=5sin beta then (tan((alpha+beta)/2))/(tan((alpha-beta)/2))=

If sin(alpha+beta)=1 and sin(alpha-beta)=1/2 where alpha,beta in [0,pi/2] , then tan(2alpha+beta)=

sin alpha + sin beta= 1/4 and cos alpha + cos beta = 1/3 The vlaue of tan (alpha + beta) is