A alyren Junior Member Joined Sep 9, 2010 Messages 59 Nov 15, 2010 #1 establish the identity [cos(x-y)]/[cos(x+y)] = [1+ tanx tany] / [1- tanx tany] got stuck in this step. (cosx cosy + sinx siny) / (cosx cosy - sinx siny)
establish the identity [cos(x-y)]/[cos(x+y)] = [1+ tanx tany] / [1- tanx tany] got stuck in this step. (cosx cosy + sinx siny) / (cosx cosy - sinx siny)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Nov 15, 2010 #2 Hello, alyren! \(\displaystyle \text{Establish the identity: }\;\frac{\cos(x-y)}{\cos(x+y)} \:=\: \frac{1+ \tan x\tan y}{1- \tan x\tan y}\) Click to expand... \(\displaystyle \text{You had: }\;\frac{\cos(x-y)}{\cos(x+y)} \;\;=\;\;\frac{\cos x\cos y + \sin x\sin y}{cos x\cos y - \sin x\sin y}\) \(\displaystyle \text{Divide numerator and denominator by }\cos x\cos y\) . . \(\displaystyle \frac{\dfrac{\cos x\cos y}{\cos x\cos y} + \dfrac{\sin x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y}{\cos x\cos y} - \dfrac{\sin x\sin y}{\cos x\cos y}} \;\;=\;\;\frac{1 + \dfrac{\sin x}{\cos x}\!\cdot\!\dfrac{\sin y}{\cos y}}{1 - \dfrac{\sin x}{\cos x}\!\cdot\!\dfrac{\sin y}{\cos y}} \;\;=\;\;\frac{1 + \tan x\tan y}{1 - \tan x\tan y}\)
Hello, alyren! \(\displaystyle \text{Establish the identity: }\;\frac{\cos(x-y)}{\cos(x+y)} \:=\: \frac{1+ \tan x\tan y}{1- \tan x\tan y}\) Click to expand... \(\displaystyle \text{You had: }\;\frac{\cos(x-y)}{\cos(x+y)} \;\;=\;\;\frac{\cos x\cos y + \sin x\sin y}{cos x\cos y - \sin x\sin y}\) \(\displaystyle \text{Divide numerator and denominator by }\cos x\cos y\) . . \(\displaystyle \frac{\dfrac{\cos x\cos y}{\cos x\cos y} + \dfrac{\sin x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y}{\cos x\cos y} - \dfrac{\sin x\sin y}{\cos x\cos y}} \;\;=\;\;\frac{1 + \dfrac{\sin x}{\cos x}\!\cdot\!\dfrac{\sin y}{\cos y}}{1 - \dfrac{\sin x}{\cos x}\!\cdot\!\dfrac{\sin y}{\cos y}} \;\;=\;\;\frac{1 + \tan x\tan y}{1 - \tan x\tan y}\)
