Let `f(x)={(tan)pi/4+tanx}{(tan)pi/4+(tan)(pi/4-x)}and g(x)=x^2+1.` Then `g{f(x)}+f'(x)=`

If f(x)=[Tanpi//4+Tanx].[Tanpi//4+tan(pi//4-x)] and g (x)=x(x+1) , then the value of g'[f(x)] is equal to

tan x+(tan((pi)/(4))+tan x)/(1-tan((pi)/(4))tan x)=2

(tan(pi/4+x))/(tan(pi/4-x)) = ((1+tanx)/(1-tanx))^(2)

(tan(pi/4+x))/(tan(pi/4-x)) = ((1+tanx)/(1-tanx))^(2)

f(x)=[Tan( pi/4)+Tan x]*[Tan (pi/4)+Tan(pi/4-x)]" and "g(x)=x(x+1)" .Then the value of "g[f(x)]" is equal to "

f(x)=tan x,x in(-(pi)/(2),(pi)/(2)) and g(x)=sqrt(1-x^(2)) then g(f(x)) is

Let f(x) = tan x, x in (-pi/2,pi/2)and g(x) = sqrt(1-x^2) then g(f(x)) is

Prove that: (tan(pi/4+x))/(tan(pi/4-x))=((1+tanx)/(1-tanx))^2

prove that (tan(pi/4+x))/(tan(pi/4-x))=((1+tanx)/(1-tanx))^2

(tan((pi)/(4)+x))/(tan((pi)/(4)-x))=((1+tan x)/(1-tan x))^(2)