cotB/(secB-tanB)-cosB/(secB+tanB)= sinB+cscBcotBsecB−tanB−cosBsecB+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+cscBcotB(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)cotB⋅secB+cotB⋅tanBsec2B−tan2B−cosB⋅secB−cosB⋅tanBsec2B−tan2B
Before we combine let's apply one Pythagorean identity:
1+tan^2x=sec^2x1+tan2x=sec2x
Manipulated to:
1=sec^2x-tan^2x1=sec2x−tan2x
That should look familiar as sec^2x-tan^2xsec2x−tan2x is our denominator, so let's substitute a 1 in there:
(cotB*secB+cotB*tanB)/(1)-(cosB*secB-cosB*tanB)/(
1)cotB⋅secB+cotB⋅tanB1−cosB⋅secB−cosB⋅tanB1
There's still one more step before we combine, we need to utilize reciprocal and quotient identities to make the expression simpler:
tanx= sinx/cosxtanx=sinxcosx
cotx= 1/tanxcotx=1tanx
secx= 1/cosxsecx=1cosx
cotx= cosx/sinxcotx=cosxsinx
1/sinx= cscx1sinx=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
