This is an identity. It is true for all values of X.
Perhaps you wish to demonstrate it is true. We use these identities.
Csc[x] = 1/Sin[x]
Sec[x] = 1/because[x]
Tan[x] = Sin[x]/because[x]
Cot[x] = because[x]/Sin[x]
Sin[x]2 + because[x]2 = 1
(Csc[x]+Sec[x])/(Sin[x]+because[x]) = Cot[x] + Tan[x]
(1/Sin[x] + 1/because[x])/(Sin[x] + because[x]) = because[x]/Sin[x] + Sin[x]/because[x]
((because[x] + Sin[x])/(Sin[x]because[x]))/(Sin[x] + (because[x]) =because[x]2+Sin[x]2)/(Sin[x]because[x])
(1/(Sin[x]because[x]))((because[x] + Sin[x])/(because[x] + Sin[x])) = 1/(Sin[x]because[x])
1/(Sin[x]because[x]) = 1/(Sin[x]because[x]) Q.E.D.
Perhaps you wish to demonstrate it is true. We use these identities.
Csc[x] = 1/Sin[x]
Sec[x] = 1/because[x]
Tan[x] = Sin[x]/because[x]
Cot[x] = because[x]/Sin[x]
Sin[x]2 + because[x]2 = 1
(Csc[x]+Sec[x])/(Sin[x]+because[x]) = Cot[x] + Tan[x]
(1/Sin[x] + 1/because[x])/(Sin[x] + because[x]) = because[x]/Sin[x] + Sin[x]/because[x]
((because[x] + Sin[x])/(Sin[x]because[x]))/(Sin[x] + (because[x]) =because[x]2+Sin[x]2)/(Sin[x]because[x])
(1/(Sin[x]because[x]))((because[x] + Sin[x])/(because[x] + Sin[x])) = 1/(Sin[x]because[x])
1/(Sin[x]because[x]) = 1/(Sin[x]because[x]) Q.E.D.