`tan^(-1)x+cot^(-1)x=(pi)/(2)`

If tan^(-1)x+2cot^(-1)x=(2pi)/(3) then x =

Solve for x : tan^(-1)x+2cot^(-1)x=(2pi)/3

Solve: tan^(-1)x+2cot^(-1)x=(2pi)/3

It is given that A=(tan^(-1)x)^(3)+(cot^(-1)x)^(3) where x gt 0 and B=(cos^(-1)t)^(2)+(sin^(-1)t)^(2) where t in [0, (1)/(sqrt(2))] , and sin^(-1)x+cos^(-1)x=(pi)/(2) for -1 le x le 1 and tan^(-1)x +cot^(-1)x=(pi)/(2) for x in R . The interval in which A lies is :

It is given that A=(tan^(-1)x)^(3)+(cot^(-1)x)^(3) where x gt 0 and B=(cos^(-1)t)^(2)+(sin^(-1)t)^(2) where t in [0, (1)/(sqrt(2))] , and sin^(-1)x+cos^(-1)x=(pi)/(2) for -1 le x le 1 and tan^(-1)x +cot^(-1)x=(pi)/(2) for x in R . The maximum value of B is :

If tan^(-1)x+2cot^(-1)x=(2pi)/3, then x , is equal to (a) (sqrt(3)-1)/(sqrt(3)+1) (b) 3 (c) sqrt(3) (d) sqrt(2)

If tan^(-1)x+2cot^(-1)x=(5pi)/6 , then find x.

Solve: 5tan^(-1)x+3cot^(-1)x=2pi

If tan^(-1)x+tan^(-1)y=(pi)/(4) , then cot^(-1)x+cot^(-1)y=

If cot^(-1)x+cot^(-1)y+cot^(-1)z=(pi)/(2) , then x+y+z=