If sin theta. Cos theta =k then sin theta + cos theta =

Let theta be @ for simpler in writingLet`s sin@ + cos@=m then after doing both side squaring we get 1+2k=m^2 so therefore m=(1+2k)^1/2

Let`s assume sintheta as a and costtheta as b then He given ab=k then you have to find a+b(a+b)^2-2ab=(a^2+b^2-

[Sin(tita)+cos(tita)]^2=sin^2(tita)+cos^2(tita)+2.sin(tita) .cos (tita) =1+2.kSo,sin (tita)+cos(tita) = √1+2kSince we know thatSin^2(tita)+cos^2(tita) = 1Given sin(tita).cos(tita) = k