#cotB/(secB-tanB)-cosB/(secB+tanB)= sinB+cscB#
A common denominator to add these fractions:
#(cotB(secB+tanB))/((secB-tanB)(secB+tanB))-(cosB(secB-tanB))/((secB+tanB)(secB-tanB))= sinB+cscB#
#(cotB*secB+cotB*tanB)/(sec^2B-tan^2B)-(cosB*secB-cosB*tanB)/(sec^2B-tan^2B)#
Before we combine let's apply one Pythagorean identity:
#1+tan^2x=sec^2x#
Manipulated to:
#1=sec^2x-tan^2x#
That should look familiar as #sec^2x-tan^2x# is our denominator, so let's substitute a 1 in there:
#(cotB*secB+cotB*tanB)/(1)-(cosB*secB-cosB*tanB)/(
1)#
There's still one more step before we combine, we need to utilize reciprocal and quotient identities to make the expression simpler:
#tanx= sinx/cosx#
#cotx= 1/tanx#
#secx= 1/cosx#
#cotx= cosx/sinx#
#1/sinx= cscx#
So:
#(cancel(cosB)/sinB*1/cancel(cosB)+1/cancel(tanB)*cancel(tanB))/(1)-(cancel(cosB)*1/cancel(cosB)-cancel(cosB)*sinB/cancel(cosB))/(
1)#
Leaves us with:
#cscB+1-(1-sinB)#
Distribute the negative:
#cscBcancel(+1)cancel(-1)+sinB#
Gets us to:
#sinB+cscB#
In case confusion is caused from where #cscB# came from: remember: #1/sinx= cscx#
