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]

(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]

(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.