tas tan(x)={[(x^(2)-3x+2)/(x-3)," for "0<=x<4],[(x^(2)-1)/(x-2)," for "4<=x<=6]," then on "[0,6)

Show that tan 3 x=(3 tan x-tan ^(3) x)/(1-3 tan ^(2) x)

Let |{:(tan^(-1)x, tan^(-1)2x, tan^(-1)3x), (tan^(-1)3x, tan^(-1)x, tan^(-1)2x), (tan^(-1)2x, tan^(-1)3x, tan^(-1)x):}|=0 , then the number of values of x satisfying the equation is

Prove that tan 3x= (3tan x-tan^3 x)/(1-3tan^2 x)

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

If f(x)=((3x+tan^(2) x)/x) is continuous at x=0 , then f(0) is equal to.

Prove that tan^(-1) ((3x-x^(3))/(1-3x^(2)))=tan^(-1)x +"tan"^(-1)(2x)/(1-x^(2)), |x| lt (1)/(sqrt(3)) .

Solve 3tan2x-4tan3x=tan^(2)3x tan2x

lim_(x rarr 0) ((sin^(-1)3x)^(3) . tan x)/((tan^(-1)x)^(2).x^(2)) =

tan^(- 1)x+tan^(- 1)\ (2x)/(1-x^2)=pi+tan^(- 1)\ (3x-x^3)/(1-3x^2),(x >0) is true if