Question

f(sinx)=1+(tgx)^2, what is f(1/sinx)

Answer 1

I've never seen (tgx) before. What does it mean?

Answer 2

oh, tan(x), I see.

Answer 3

Interesting I have never seen that notation either. I googled it though and it's supposed to be "tan x" (tangent of x )

Answer 4

Well you could start by simplifying f(sinx)=1+(tgx)^2=1+(sinx)^2/(cosx)^2=((cosx)^2+(sinx)^2)/(cosx)^2 = 1/(cosx)^2 = 1/(1-(sinx)^2). Then replace sinx by 1/sinx: f(1/sinx) = 1/(1-(1/sinx)^2).

Answer 5

You can rewrite the last expression into something somewhat simpler: (sinx)^2/((sinx)^2-1)=(sinx)^2/(cosx)^2=(tgx)^2. Interesting exercise!

Answer 6

My last post was vague and wrong. I'll try again.

f(1/sinx)=1/(1-(1/sinx)^2)=1/(1-(1/sinx)^2)*(sinx)^2/(sinx)^2=(sinx)^2/((sinx)^2-1)=(sinx)^2/((sinx)^2-(cosx)^2-(sinx)^2)=-(sinx)^2/(cosx)^2=-(tgx)^2

Answer 7

Thank you, it took me a while until I got all that but I understand now. You helped a lot.

Answer 8

You're welcome.

I want to add a thing which I just noticed. Actually, we can't know for sure what f(1/sinx) is. f(sinx)=1+(tgx)^2 only tells us about the values of f(y) when -1 < y < 1 because that's the range of sinx. 1/sinx is never in that range so we know nothing about f(1/sinx).

However, one could argue that my solution is the "simplest".

The thing that went wrong in my reasoning was that I replaced sinx by 1/sinx without considering their ranges.