If `f(x)=tanx+tan^2 x tan 2x and g(n)=sum_(r=0)^n f(2^r) and l=g(2015)-tan(2^2016)`, then

If f(x)=tan x+tan^(2)x tan2x and g(n)=sum_(r=0)^(n)f(2^(r)) and l=g(2015)-tan(2^(201) then

If f(x)=tan x+tan^(2)x tan2x and g(n)=sum_(r=0)^(x=k)f(2^(r)) then g(2013)-tan(2^(2014)) is equal to ( )0(B) tan 1(C)-tan1(D)1

tanx + tan2x + tan3x = tanx tan2x tan3x

f(x)=4tanx-tan^(2)x+tan^(3)x,xnenpi+(pi)/(2) ,

If f(x) = tan^(-1)((2^x)/(1 + 2^(2x + 1))) , then sum_(r = 0)^(9) f^(r ) is

If f(x) = tan^(-1)((2^x)/(1 + 2^(2x + 1))) , then sum_(r = 0)^(9) f(r ) is

Differentiate (tanx tan 2x tan 3x tan 4x) w.r.t. x.

Show that tanx tan 2x tan 3x = tan 3x-tan 2x- tan x

If f(x) = sin x tan x and g(x) = x^(2) then in interval x in (0, pi//2) is

If f(x) = lim_(n->oo) sum_(r=0)^n (tan(x/2^(r+1)) + tan^3 (x/2^(r+1)))/(1- tan^2 (x/2^(r+1))) then lim_(x->0) f(x)/x is