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

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

sin(alpha+beta)=1 and sin(alpha-beta)=(1)/(2)quad 0<=alpha,beta,<=(pi)/(2), then find tan(alpha+2 beta) and tan(2 alpha+beta)

csc^(2)(alpha+beta)-sin^(2)(beta-alpha)+sin^(2)(2 alpha-beta)=cos^(2)(alpha-beta) where alpha,beta in(0,(pi)/(2)), then sin(alpha-beta) is equals

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

sin^(4)alpha+4cos^(4)beta+2=4sqrt(2)sin alpha cos beta;alpha,beta in[0,pi], then cos(alpha+beta)-cos(alpha-beta) is equal to :

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

If : sin (alpha + beta)=1 and sin (alpha-beta)=(1)/(2), "then" : tan (alpha+2beta)*tan(2alpha+beta)= A)-1 B)0 C)1 D)2