a function with the property f(1+2)=f(1)+f(2)/1-f(1)f(2) is?

Question

f(x)=tanx
f(x)=sinx
f(x)=secx
f(x)=e^x
f(x)=-1

Answer 1

f(x)=tanx

Becase tan(A+B)=(tanA+tanB)/(1-tanA.tanB)

Answer 2

Notation: i'm making use of _ as a subscript indicator to designate words contained interior the sequence. a) (fn,fn+a million)= a million i think the () consequently is meant to prepare the main nicely known common factor. think of that for some value of n, (f_n, f_(n+a million)) = ok > a million it particularly is, some pair of successive words contained interior the sequence have an common factor greater advantageous than a million. Then pondering the undeniable fact that ok divides the two f_n and f_(n+a million), it of course additionally divides f_(n-a million) = f_(n+a million) - f_n and it additionally divides f_(n+2) = f_n + f_(n+a million) and an identical reasoning helps us to spice up divisibility by using making use of ok to all words contained interior the sequence. yet it style of feels that the 1st 2 words are no extra divisible by using making use of ok if ok>a million, so our supposition that 2 successive words have an common factor greater advantageous than a million leads to a contradiction and could be pretend. for this reason, for any n, (f_n, f_(n+a million)) = a million b) (f_(n+a million))^2= f_n*f_(n+2) + (-a million)^n think of this formula holds for some n=ok; it particularly is, [f_(ok+a million)]^2 = f_k f_(ok+2) + (-a million)^ok Then (f_(ok+2))^2 = [f_(ok+a million) + f_k] [f_(ok+3) - f_(ok+a million)] = f_(ok+a million) f_(ok+3) - [f_(ok+a million)]^2 + f_k f_(ok+3) - f_k f_(ok+a million) = f_(ok+a million) f_(ok+3) - [f_(ok+a million)]^2 + f_k [f_(ok+2) + f_(ok+a million)] - f_k f_(ok+a million) = f_(ok+a million) f_(ok+3) - [f_(ok+a million)]^2 + f_k f_(ok+2) + f_k f_(ok+a million) - f_k f_(ok+a million) = f_(ok+a million) f_(ok+3) - [f_(ok+a million)]^2 + f_k f_(ok+2) = f_(ok+a million) f_(ok+3) - [f_k f_(ok+2) + (-a million)^ok] + f_k f_(ok+2) = f_(ok+a million) f_(ok+3) - f_k f_(ok+2) + (-a million)^ok + f_k f_(ok+2) - (-a million)^ok = f_(ok+a million) f_(ok+3) + (-a million)(-a million)^ok = f_(ok+a million) f_(ok+3) + (-a million)^(ok+a million) it particularly is an same formula for n=ok+a million. pondering the undeniable fact that we are in a position to truly reveal that the formula holds for n=a million: a million^2 = a million*2 + (-a million)^a million the formula holds for all integers n>a million by using making use of mathematical induction.

Answer 3

f(x)=tanx



because tan(1+2)=tan1+tan2/1-tan1*tan2

Answer 4

f(x)=tanx

Answer 5

tan(a+b)=tan(a)+tan(b)/1-tan(a).tan(b)