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

#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...

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