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

Write the value of tan^(-1)a/b-tan^(-1)((a-b)/(a+b))

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

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

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=

The value of sec[tan^(-1)((b+a)/(b-a))-tan^(-1)((a)/(b))] is

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

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)