cot/1-tan + tan/1-cot = 1+tan+cot
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Nov 19, 2010 #2 Hello, kowens! \(\displaystyle \frac{\cot x}{1-\tan x} + \frac{\tan x}{1-\cot x} \:=\: 1+\tan x+\cot x\) Click to expand... \(\displaystyle \displaystyle \text{The left side is: }\;\frac{\dfrac{\cos x}{\sin x}}{1 - \dfrac{\sin x}{\cos x}} + \frac{\dfrac{\sin x}{\cos x}}{1 - \dfrac{\cos x}{\sin x}}\) \(\displaystyle \text{Multiply the fractions by }\frac{\sin x\cos x}{\sin x\cos x}\!:\quad\frac{\cos^2\!x}{\sin x\cos x-\sin^2x} \;+\; \frac{\sin^2\!x}{\sin x\cos x - \cos^2\!x} \;\;=\;\;\frac{\cos^2\!x}{\sin x(\cos x - \sin x)} \;+\; \frac{\sin^2\!x}{-\cos x(\cos x - \sin x)}\) \(\displaystyle \text{Get a common denominator: }\;\frac{\cos^3\!x}{\sin x\cos x(\cos x-\sin x)} \;-\; \frac{\sin^3\!x}{\sin x\cos x(\cos x - \sin x)} \;\;=\;\;\frac{\cos^3\!x - \sin^3\!x}{\sin x\cos x(\cos x - \sin x)}\) \(\displaystyle \text{Factor and reduce: }\;\frac{(\cos x - \sin x)(\cos^2\!x + \sin x\cos x + \sin^2\!x)}{\sin x\cos x(\cos x - \sin x)} \;\;=\;\; \frac{\cos^2\!x + \sin x\cos x + \sin^2\!x}{\sin x\cos x} \;\;=\;\;\frac{\sin x\cos x + \sin^2\!x + \cos^2\!x}{\sin x\cos x}\) \(\displaystyle \text{And we have: }\;\frac{\sin x\cos x}{\sin x\cos x} \;+\; \frac{\sin^2\!x}{\sin x\cos x} \;+\; \frac{\cos^2\!x}{\sin x\cos x} \;\;=\;\; 1 \;+\; \frac{\sin x}{\cos x} \;+\; \frac{\cos x}{\sin x} \;\;=\;\;1 \;+\; \tan x \;+\; \cot x\)
Hello, kowens! \(\displaystyle \frac{\cot x}{1-\tan x} + \frac{\tan x}{1-\cot x} \:=\: 1+\tan x+\cot x\) Click to expand... \(\displaystyle \displaystyle \text{The left side is: }\;\frac{\dfrac{\cos x}{\sin x}}{1 - \dfrac{\sin x}{\cos x}} + \frac{\dfrac{\sin x}{\cos x}}{1 - \dfrac{\cos x}{\sin x}}\) \(\displaystyle \text{Multiply the fractions by }\frac{\sin x\cos x}{\sin x\cos x}\!:\quad\frac{\cos^2\!x}{\sin x\cos x-\sin^2x} \;+\; \frac{\sin^2\!x}{\sin x\cos x - \cos^2\!x} \;\;=\;\;\frac{\cos^2\!x}{\sin x(\cos x - \sin x)} \;+\; \frac{\sin^2\!x}{-\cos x(\cos x - \sin x)}\) \(\displaystyle \text{Get a common denominator: }\;\frac{\cos^3\!x}{\sin x\cos x(\cos x-\sin x)} \;-\; \frac{\sin^3\!x}{\sin x\cos x(\cos x - \sin x)} \;\;=\;\;\frac{\cos^3\!x - \sin^3\!x}{\sin x\cos x(\cos x - \sin x)}\) \(\displaystyle \text{Factor and reduce: }\;\frac{(\cos x - \sin x)(\cos^2\!x + \sin x\cos x + \sin^2\!x)}{\sin x\cos x(\cos x - \sin x)} \;\;=\;\; \frac{\cos^2\!x + \sin x\cos x + \sin^2\!x}{\sin x\cos x} \;\;=\;\;\frac{\sin x\cos x + \sin^2\!x + \cos^2\!x}{\sin x\cos x}\) \(\displaystyle \text{And we have: }\;\frac{\sin x\cos x}{\sin x\cos x} \;+\; \frac{\sin^2\!x}{\sin x\cos x} \;+\; \frac{\cos^2\!x}{\sin x\cos x} \;\;=\;\; 1 \;+\; \frac{\sin x}{\cos x} \;+\; \frac{\cos x}{\sin x} \;\;=\;\;1 \;+\; \tan x \;+\; \cot x\)
