`sin(x+y)=1, cos(x-y)=sqrt(3)/2` and `x>y`,`0^@<= x+y<=90^@`, then find `x` and `y`.

Solve the equation :sin(x-y)=(sqrt(3))/(2) and cos(x+y)=(1)/(2)

If cos (x-y)=(sqrt3)/2 and sin(x+y)=1/2 , then the value of x(0ltxlt90) is

cos (x + y), sin (x + y), - cos (x + y) sin (xy), cos (xy), sin (xy) sin2x, 0, sin2y] | = sin2 (x + y)

if x+y=(2 pi)/(3) and cos x+cos y=(sqrt(3))/(2) then x,y is equal to:

If (sin x)/(sin y)=(1)/(2),(cos x)/(cos y)=(3)/(2), where x,y,in(0,(pi)/(2)), then the value of tan(x+y) is equal to (a)sqrt(13)(b)sqrt(14)(c)sqrt(17)(d)sqrt(15)

Let f(x,y)=sin(2x-y)cos y+cos(2x-y)sin y for all x,y in R. The value of lim_(x rarr0)(1+tan x-x)^((1)/(x^(2)f(x,y))) is equal to

If sin(x-y)-cos(x+y)=(1)/(2) then the values of x&y lying between 0 and pi are given by

If (sin x)/(sin y)=(1)/(2),(cos x)/(cos y)=(3)/(2), where x,y in(0,(pi)/(2)), then the value of tan(x+y) is equal to

For 0