Question

How do you prove `[sin(x+y) / sin(x-y)] = [(tan(x) + tan (y) )/ (tan (x) - tan (y))]`?

Answer 1

Please refer to a Proof in the Explanation.

Explanation:

We have, `tanx+tany=sinx/cosx+siny/cosy`,

`=(sinxcosy+cosxsiny)/(cosxcosy)`,

`rArr tanx+tany=sin(x+y)/(cosxcosy)...................(square^1)`.

Similarly, `tanx-tany=sin(x-y)/(cosxcosy)............(square^2)`.

`:." by "(square^1) and (square^2), (tanx+tany)/(tanx-tany)`,

`=sin(x+y)/sin(x-y)`.

Hence, the Proof.

Answer 2

We seek to prove that:

`sin(x+y)/sin(x-y) -= (tanx+tany)/(tanx-tany)`

We can the trigonometric identities:

`sin(A+B) -= sinAcosB + cosAsinB`
`sin(A-B) -= sinAcosB - cosAsinB`

Consider the LHS:

`LHS = sin(x+y)/sin(x-y)`

`\ \ \ \ \ \ \ \ = (sinxcosy + cosxsiny)/(sinxcosy - cosxsiny)`

Now if we multiply both numerator and denominator by `1/(cosxcosy)` we get:

`LHS = (sinxcosy + cosxsiny)/(sinxcosy - cosxsiny) * (1/(cosxcosy)) / (1/(cosxcosy))`

`\ \ \ \ \ \ \ \ = ( (sinxcosy)/(cosxcosy) + (cosxsiny)/(cosxcosy))/ ((sinxcosy)/(cosxcosy) - (cosxsiny)/(cosxcosy))`

`\ \ \ \ \ \ \ \ = ( (sinx)/(cosx) + (siny)/(cosy))/ ((sinx)/(cosx) - (siny)/(cosy))`

`\ \ \ \ \ \ \ \ = ( tanx + tany )/ ( tanx - tany ) \ \ \ ` QED