If `tanx=a/(a+1)` and `tany=1/(2a+1)` then show that one of the values of `x+y=pi/4; a in R, a!= -1, a!= -1/2`

If tan x= a/(a+1) and tan y=1/(2a+1) , show that one of the values of : x+y= pi/4, a in R, a ne -1, a ne -1/2 .

If x+y=pi/4 and tanx+tany=1, then

If x+y =pi/4 and tanx + tany=1 , then

If tanx=1/7,tany=1/3 ,then x+y is:

If x + y + z = pi , tanx tany = 2 tanx + tany + tanz = 6 , then z equals :

If y=logsqrt(tanx) , then the value of (dy)/(dx) at x=pi/4 is given by (a) oo (b) 1 (c) 0 (d) 1/2

Solve: tanx + tany =1, x+y=pi/4

If 1-tanx/1+tanx=tany and x-y=pi/6 , then x, y are respectively

If tan^(-1)x+tan^(-1)y=(pi)/(4) , and the value of xy<1 ,then show that x+y+xy=1 .

If tanx tany=a and x+y=2b show that tanx and tany are the roots of the equation z^2-(1-a)tan2b*z+a=0