Question

(sinxcosy + cosxsiny)/ (cosxcosy - sinxsiny) = (tanx+tany)/(1-tanxtany)

SIMPLIFY

Answer 1

pls help! need step by step explanation please?

Answer 2

All I did was use the identity

\(\tan \theta = \dfrac{\sin \theta}{\cos \theta} \)

on the right side for all tangents.

Answer 3

Do you follow so far?

Answer 4

YES

Answer 5

Great.
All you need is one more step.

Answer 6

On the right you have tan(x-y)

Answer 7

multiply by reciprocal?

Answer 8

yea theres a right side

Answer 9

On the left you have sin(x+y)/cos(x+y)

Answer 10

Multiply the right fraction by

\(\dfrac{ \cos x \cos y}{\cos x \cos y}\)

Answer 11

So both sides are equal to tan(x+y) in 2 easy steps.

Answer 12

so then what would it look like

Answer 13

but u said u multiply the tan (x-y) by that

Answer 14

\[\frac{sinxcosy+cosxsiny}{cosxcoxy-sinxsiny}=\frac{\sin(x+y)}{\cos(x+y)}=\tan(x+y)\]

Answer 15

thats it?

Answer 16

RS:
\[\frac{\tan x+\tan y}{1-\tan x \tan y}=\tan (x+y)\]

Answer 17

ohg so its like a proof

Answer 18

cuz sin/cos=tan

Answer 19

\[\tan (x+y)=\tan (x+y)\]

Answer 20

ok thank you both!

Answer 21

what if u had a question like:
sec^2Θ= 1 - tanΘ??? how would you simply it then

Answer 22

Is that supposed to be tan^2 on the right?

Answer 23

no just tan theta

Answer 24

Is that an identity you are supposed to prove or what?