Show that `(tan^-1)(2ab)/(a^2-b^2)+(tan^-1)(2xy)/(x^2-y^2)=(tan^-1) (2alphabeta)/(alpha^2-beta^2),` where `alpha=ax- by and beta=ay+bx.`

For any a,b,c,x,y>0 prove that :(2)/(3)tan^(-1)((3ab^(2)-a^(3))/(b^(3)-3a^(2)b))+(2)/(3)tan^(-1)((3xy^(2)-x^(3))/(y^(3)-3x^(2)y))=tan^(-1)(2 alpha beta)/(a^(2)-beta^(2)) where alpha=-ax+by,beta=bx+ay

(1+tan alpha tan beta)^2 + (tan alpha - tan beta)^2 =

(1+tan alpha tan beta)^(2)+(tan alpha-tan beta)^(2)=

(1+tan alpha tan beta)^(2) + (tan alpha -tan beta)^(2) =

If tan^(-1)(alpha+i beta)=a+ib, then a= 1). (1)/(2)tan^(-1)((2 alpha)/(1-alpha^(2)-beta^(2))) 2) . (1)/(2)tan^(-1)((2 alpha)/(1+alpha^(2)+beta^(2))) 3). tan^(-1)((2 alpha)/(1-alpha^(2)-beta^(2))) 4). (1)/(2)tan^(-1)((2 beta)/(1+alpha^(2)+beta^(2)))

Prove that: tan^(-1){(cos2 alpha sec2 beta+cos2 beta sec2 alpha)/(2)}=tan^(-1){tan^(2)(alpha+beta)tan^(2)(alpha-beta)}+tan

Prove that: tan^(-1){(cos2alphasec2beta+cos2betasec2alpha)/2}=tan^(-1){tan^2(alpha+beta)tan^2(alpha-beta)}+tan^(-1)1

If sin(alpha+beta)=1,sin(alpha-beta)=(1)/(2) then tan(alpha+2 beta)tan(2 alpha+beta)=