cot(x=y)=cotx coty-1/cotx+coty?

Question

prove the identity

CORRECTION!

cot(x+y)=cotx coty-1/cotx+coty

Answer 1

plz write the '+' sign correctly
cot(x+y)=1/tan(x+y)
tan(x+y)=(tanx+tany)/
(1-tanxtany)
cot(x+y)=(1-tanxtany)/
(tanx+tany)
= 1/(tanx+tany) - tanxtany/(tanx+tany)
[tanx+tany= 1/cotx +1/coty
= (coty+cotx)/cotxcoty
tanxtany= 1/cotxcoty]
cot(x+y)= cotxcoty/(cotx+coty) -
[(cotxcoty)(1/cotxcoty)]/
(cotx+coty)
= [cotxcoty-1]/(cotx+coty)

Answer 2

cot (x+y)
= 1 / tan (x+y)
= 1/ [(tan x + tan y)/(1 - tan x tan y)]
= (1 - tan x tan y) / (tan x + tan y)
= (1 - (1/(cot x cot y))) / (1/cot x + 1/cot y)
= ((cot x cot y - 1)/(cot x cot y)) / ((cot x + cot y)/(cot x cot y))
= (cot x cot y - 1) / (cot x + cot y)
Proved !!!

Answer 3

I think you mean cot(x+y)=(cot(x)cot(y)-1)/(cot(x)+cot(y))

start with the definitions

cot(x+y)=cos(x+y)/sin(x+y)=

(cos(x)cos(y)-sin(x)sin(y))/(sin(x)cos(y)+sin(y)cos(x))

pull out sin(x)sin(y) from numerator and denominator
and cancel and you'll get this intermediate step

((cos(x)cos(y))/(sin(x)sin(y))-1)/(cos(y)/sin(y)-cos(x)/cos(y))

yielding

(cot(x)cot(y)-1)/(cot(x)+cot(y)) QED