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)=`

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

Solve: tan(pi/4+x)+tan(pi/4-x)=2

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)

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

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

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

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