How do you prove {(sin(x + y)) / sin(x)cos(y)} = 1-cot(x)tan(y){sin(x+y)sin(x)cos(y)}=1cot(x)tan(y)?

2 Answers
Apr 7, 2017

Q.E.D. See below.

Explanation:

I'm going to assume you meant sin(x+y)/(sinxcosy)=1+cotxtanysin(x+y)sinxcosy=1+cotxtany and just forgot parentheses on the left side denominator and the plus on the right side, otherwise the identity is false.

Let's start working the left hand side by using the sine addition formula sin(alpha+beta)=sinalphacosbeta+cosalphasinbetasin(α+β)=sinαcosβ+cosαsinβ
sin(x+y)/(sinxcosy)=(sinxcosy+cosxsiny)/(sinxcosy)sin(x+y)sinxcosy=sinxcosy+cosxsinysinxcosy

Then split the numerator over the denominator
(sinxcosy+cosxsiny)/(sinxcosy)=(sinxcosy)/(sinxcosy)+(cosxsiny)/(sinxcosy)sinxcosy+cosxsinysinxcosy=sinxcosysinxcosy+cosxsinysinxcosy

Simplify and factor
(sinxcosy)/(sinxcosy)+(cosxsiny)/(sinxcosy)=1+(cosx/sinx)(siny/cosy)sinxcosysinxcosy+cosxsinysinxcosy=1+(cosxsinx)(sinycosy)

And knowing that tantheta=sintheta/costhetatanθ=sinθcosθ and cottheta=costheta/sinthetacotθ=cosθsinθ,
1+(cosx/sinx)(siny/cosy)=1+cotxtany1+(cosxsinx)(sinycosy)=1+cotxtany

Therefore the left side is equal to the right side. Q.E.D.

Apr 7, 2017

Assuming the questioner meant (sin(x+y))/(sinx cosy)=1+cotx tanysin(x+y)sinxcosy=1+cotxtany, then the proof is as follows:

Explanation:

(sin(x+y))/(sinx cosy) = (sinxcosy+cosxsiny)/(sinx cosy)=sin(x+y)sinxcosy=sinxcosy+cosxsinysinxcosy=

(sinxcosy)/(sinxcosy) + (cosx)/sinx xx siny/cosy=sinxcosysinxcosy+cosxsinx×sinycosy=

1+cotx tany1+cotxtany QED/ΟΕΔ