(sinx +sinx/cosx)/1+secx

by taking L.C.M

(sinxcosx+sinx)/1+1/cosx

sinx(cosx +1) / (cosx+1)/cosx

then (cosx+1) and(cosx+1) and cosx and cosx will get cancel

and remaining after cancelation will be sinx.

(sinx + tanx)/(1 +secx)= sinx

you can also consider

multiplying both sides by your denominator

(1 +secx) and get

sinx + tanx = sinx +sinxsecx

use the identity for secx = 1/cosx

sinx + tanx= sinx + sinx(1/cosx)

sinx + tanx= sinx + sinx/cosx

tanx=sinx/cosx

sinx + tanx= sinx + tanx

(sinx+sinx/cosx) / (1+1/cosx)=sinx

change tangent and secant to sinx/cosx and 1/cosx then the simplified form of this eq. is

(sinx cos^2x+sinx cosx) / (cos^2x +cosx)

factor out the eq. above and you’ll get

sinx cosx(cosx + 1) / cosx (1+cos)

cancel the cosx and 1+cosx and the only term that will be left is sinx

so

sinx=sinx