Precalculus...Verifying trigonometric identities?

Question

How would I design a 2-column proof to verify this trigonometric identity?
(I can only use material up to pythagorean identities to solve this...such as sin2x+cos2x=1)

tan(x)+tan(y)
----------------------
1-[tan(x)*tan(y)]


=

cot(x) +cot(y)
-------------------
[cot(x)cot(y)] -1



(It should be in two columns on the same line but this thing deletes all the spaces and makes it look unreadable.)

Answer 1

very simple dear..
consider left hand side term
1 + 1
------- ---------
cot(x) cot(y)
= ---------------------------
1 1
1 - ----- * ----------
cot(x) cot(y)

cot(y) + cot(x)
---------------
cot(x).cot(y)
= ---------------------------------------
cot(x) * cot(y) - 1
----------------
cot(x) * cot(y)

by cancelling the denominators

= cot(y) + cot(x)
---------------------
cot(x) * cot(y) - 1

proved.

Answer 2

sin t cot t + tan t = sec t = sin t cos t / sin t + sin t / cos t = cos t + sin t / cos t = ( cos^2 t + sin t ) / cos t its not possible to be equal check your identities

Answer 3

tan(x)+tan(y)/(1-tan(x)*tan(y)
=sin(x)/cos(x) + sin(y)/cos(y) / (1-sin(x)sin(y)/cos(x)cos(y))
=[sin(x)cos(y)+cos(x)sin(y)/ cos(x)cos(y)] / [(cos(x)cos(y)-sin(x)sin(y)] / cos(x)cos(y)]
cos(x)cos*y) cancels out
=[sin(x)cos(y)+cos(x)sin(y)]/ (cos(x)cos(y)-sin(x)sin(y)]
=sin(x+y)/cos(x+y)
=tan(x+y)
The other problem is similar.

Answer 4

sin(x + y) """""""sin x cos y + cos x sin y
----------- = ----------------------------------
cos (x + y) """"""cos x cos y - sin x sin y

tan x + tan y
= --------------------
1 - tan x tan y