Start on the left side.−cot(t)+sin(t)1−cos(t)-cott+sint1-costMultiply sin(t)1−cos(t)sint1-cost by 1+cos(t)1+cos(t)1+cost1+cost.−cot(t)+sin(t)1−cos(t)1+cos(t)1+cos(t)-cott+sint1-cost1+cost1+costCombine.−cot(t)+sin(t)(1+cos(t))(1−cos(t−cot(t)+sin(t)+sin(t)cos(t)(1−cos(t))(1+cos(t))-cott+sint+sintcost1-cost1+cost))(1+cos(t))−
cot(t)+sin(t)+sin(t)cos(t)1−cos2(t)-cott+sint+sintcost1-cos2tApply pythagorean identity.−cot(t)+sin(t)+sin(t)cos(t)sin2(t)-cott+sint+sintcostsin2tWrite cot(t)cott in sines and cosines using the quotient identity.−cos(t)sin(t)+sin(t)+sin(t)cos(t)sin2(t)-costsint+sint+sintcostsin2tSimplify.1sin(t)1sintRewrite 1sin(t)1sint as csc(t)csct.csc(t)csctBecause the two sides have been shown to be equivalent, the equation is an identity.−cot(t)+sin(t)1−cos(t)=csc(t)-cott+sint1-cost=csct is an identity
