Here's the question you clicked on:
andu1854
Verify the Identity: (tan x + tan y)/(cot x + cot y) = (tan x tan y - 1 ) / (1-cot x cot y )
lhs= tanx+ tany/cotx+coty=tanx.tany rhs= tanx.tany-1/1-cotx.coty=tanx.tany
Thank you, for the help, however, I am still trying to solve it... the goal is to show the steps how the left side will equal the right side. I decided to convert tan to cos/sin and got stuck. Attempted to use tan and cot and got even more stuck. I am interested to see what I am doing wrong... Here is a PDF of what I attempted.. and the equation... I need to verify... [(tanx+tany)/(\cot x+\cot y) = (\tan x \tan y-1) / (1 -\cot x \cot y\]
@andu1854 your second step is wrong
yes, I actually started over and was able to get tan x tan y =, but I was not sure how to set up the double angle for tan, unlike for sin & cos...