Question

Prove that (1-tanx)/(1-cotx) = -tanx

I've got the first step, that (1-sinx/cosx)/(1-cosx/sinx) = -tanx, but I'm not sure what to do next. I've been told that the next step is "(cosx-sinx)/cosx * (sinx)/ (sinx - cosx) " but I'm not sure what was done to get there.

Answer 1

If it were me, I'd start differently. Pull a factor of tan x out pf the numerator to get

tan x * (1/tan x -1)/(1 - cot x)

and then use cot x = 1/ tan x.

Answer 2

Sorry, I'm a little confused about what you did there - how does (1-tanx) equal tanx*(1/tanx-1)?

Answer 3

\[1 - \tan x = (1- \tan x ) * {\tan x \over \tan x } = ({1 \over \tan x} -1)* \tan x\]

Answer 4

though fwzbangs way is faster anyway, here's what you wanted:\[{1-\tan x\over1-\cot x}={\cos x\sin x\over\cos x\sin x}\cdot{{1-\frac{\sin x}{\cos x}}\over1-\frac{\cos x}{\sin x}}={\sin x\cos x-\sin^2x\over\sin x\cos x-\cos^2x}\]\[{\sin x\over\cos x}\cdot{\cos x-\sin x\over\sin x-\cos x}\]

Answer 5

so the trick was to multiply by\[\sin x\cos x\over\sin x\cos x\]

Answer 6

Thank you both, I've got it now. I'll give best response to TuringTest for answering the original question but I really appreciate the explanation, fwizbang!