`tan^(-1)((a)/(b))-tan((a-b)/(a+b))=(pi)/(4)`

In a ABC, if C is a right angle,then tan^(-1)((a)/(b+c))+tan^(-1)((b)/(c+a))=(pi)/(3)(b)(pi)/(4)( c) (5 pi)/(2)( d) (pi)/(6)

tan^(-1)((a)/(x))+tan^(-1)((b)/(x))=(pi)/(2) then x=

if a>b>0 then the value of tan^(-1)((a)/(b))+tan^(-1)((a+b)/(a-b)) is

Prove that: tan[(pi)/(4)+(1)/(2) tan^(-1)((a)/(b))]+tan[(pi)/(4)-(1)/(2)tan^(-1)((a)/(b))]=(2sqrt(a^(2)+b^(2)))/(b)

If in triangle ABC, C = 90^circ then prove that tan^(-1) (a/(c+b)) + tan^(-1) (b/(a+c)) = pi/4 .

tan ^ (- 1) ((a) / (b)) - tan ^ (- 1) ((ab) / (a + b)) = (pi) / (4)

tan^(-1)((x)/(y))-tan^(-1)((x-y)/(x+y)) is (A) (pi)/(2)(B)(pi)/(3)(C)(pi)/(4)(D)(pi)/(4) or 3(pi)/(4) is (A) (pi)/(2)(B)

Prove that tan((pi)/(4)+(1)/(2)cos^(-1)((a)/(b)))+tan((pi)/(4)-(1)/(2)cos^(-1)((a)/(b)))=(2b)/(a)

tan^(-1)a+tan^(-1)b=(pi)/(2)

tan((pi)/(4) +(1)/(2)"cos"^(-1)(a)/(b))+ tan((pi)/(4) - (1)/(2)"cos"^(-1)(a)/(b)) is :