Question

tanx + tany / cotx + cot y = (tanx)(tany)

Make LS = RS

Answer 1

like a first step hope you know that tan x = 1/cot x

yes ? so try using this property of tan by cot

Answer 2

so after that LS would look like this? 1/cot x + 1/coty / cotx + cot y? sorry if that is messy or jumbled up

Answer 3

hmmm what terms does the fraction have? all 4 or just 2 of those ?

Answer 4

so yes this is true but first if you check the right side we need getting all terms in tan and not in cot

Answer 5

alright so it would be like this then
tanx + tany / 1/tanx + 1/tany = tanxtany?

Answer 6

exactly

Answer 7

alright what is the next step?

Answer 8

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

so hope you know how you can adding two fractions ,yes ?

Answer 9

you make a common denominator? so

tany + tan x
-----------
tanxtany?

Answer 10

yes right

Answer 11

alright i got that

Answer 12

so was hard , complicate ?

Answer 13

or easy ?

Answer 14

well that makes sense, i originally converted all terms to tan, but got a lil confused when you mentioned cot but i realised what you meant, but now what do i do with these fractions?

Answer 15

so if you add these fractions from denominator what you will get ?

Answer 16

I'm not sure what you mean

Answer 17

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

so check this please

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

this is the denominator ,yes ?

Answer 18

and what you have added so what have got ?

Answer 19

wouldnt the denominator be

tanx + tany
---------
tany + tanx
----------
tanxtany

Answer 20

thats the whole LS

Answer 21

I just forget howto work out the multiple fractions at once

Answer 22

\(\bf \cfrac{tan(x)+tan(y)}{cot(x)+cot(y)}\implies \cfrac{tan(x)+tan(y)}{\frac{1}{tan(x)}+\frac{1}{tan(y)}}\implies
\cfrac{tan(x)+tan(y)}{\frac{\bbox[border: 1px solid black]{ ? } }{tan(x)tan(y)}}\)

Answer 23

what would you get using that LCD of tan(x)tan(y) ?

Answer 24

wouldnt it be tan(y) + tan(x) where that ? is

Answer 25

or 2?

Answer 26

hmmm well, use the lcd, what would it give you? 2 or tan(y) + tan(x) ?

Answer 27

tan(x) + tan (y) because you have to do what you did to the denominator to the as well numerator right?

Answer 28

hm.... what do you mean?

Answer 29

nvm i think i just confused myself there but would it not be tany + tanx?

Answer 30

well... let's do the 1st term.... our lcd is tan(x)tan(y) our numerator is 1.... so \(\bf \cfrac{1}{tan(x)}+\frac{1}{tan(y)}\implies \cfrac{\square ?}{tan(x)tan(y)} \)

Answer 31

tanx + tany

Answer 32

tell me if im wrong lol

Answer 33

well... I wonder how you got that though?

Answer 34

anyhow, in case you forgot some about fraction sum, you divide the LCD by the denominators and the quotient is multiplied by the numerator

Answer 35

it's not tanx + tany though

Answer 36

well you have
1 1
----- + -----You have to multiple the denominator to get LCD, so left side numerator
tanx tany would multiply by tany and right side numerator would multiply by tanx

tany + tanx
----------
tanxtany

Answer 37

ahhh, right. is tan(y)+tan(x) which incidentally is the same as tan(x)+tan(y) hehee anyhow

\(\bf \cfrac{tan(x)+tan(y)}{cot(x)+cot(y)}\implies \cfrac{tan(x)+tan(y)}{\frac{1}{tan(x)}+\frac{1}{tan(y)}}\implies
\cfrac{tan(x)+tan(y)}{\frac{tan(y)+tan(x)}{tan(x)tan(y)}}\\ \quad \\
\textit{recall that }\cfrac{\quad \frac{a}{b}\quad }{\frac{c}{d}}\implies \cfrac{a}{b}\cdot \cfrac{d}{c}\qquad thus\\ \quad \\
\cfrac{tan(x)+tan(y)}{\frac{tan(y)+tan(x)}{tan(x)tan(y)}}\implies


\cfrac{\cancel{tan(x)+tan(y)}}{1}\cdot
\cfrac{tan(x)tan(y)}{\cancel{tan(y)+tan(x)}}\)

Answer 38

lol ok great thanks :D thats perfect i just forgot the method to working out like fractions in fractions

Answer 39

yw