Question

How do you prove `cos(x-y)/cos(x+y)=(cot x +tan y)/( cot x- tan y)`?

Answer 1

See below...

Explanation:

`cos(x-y)/cos(x+y)`

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

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

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

`=((cosx cdot cancel(cosy))/(sinx cdot cancel(cosy)) + (cancel(sinx) cdot siny)/(cancel(sinx) cdot cosy))/((cosx cdot cancel(cosy))/(sinx cdot cancel(cosy)) - (cancel(sinx) cdot siny)/(cancel(sinx) cdot cosy))`

`=(cot x +tan y)/( cot x- tan y)`

FORMULA reference:- wiki

hope it helps...
Thank you...

Answer 2

We seek to prove that:

`cos(x-y)/cos(x+y) -= (cotx+tany)/(cotx-tany)`

We can the trigonometric identities:

`cos(A+B) -= cosAcosB - sinAsinB`
`cos(A-B) -= cosAcosB + sinAsinB`

Consider the LHS:

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

`\ \ \ \ \ \ \ \ = (cosxcosy + sinxsiny)/(cosxcosy - sinxsiny)`

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

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

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

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

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