acos (theta) +b sin (theta)= c. Prove that asin (theta)+ b cos (theta)=+-(a^2+b^2+c^2)^(1/2)

Actually I think that the question is wrongly typed.It should probably go this way I think:acos⁡θ+bsin⁡θ=c, prove that a sin⁡θ-b cos⁡θ=±(a^2+b^2-c^2 )^(1/2)If this is the desired question, then the solution is as follows: a cos⁡θ+b sin⁡θ=csquaring both sides,weget(a cos⁡θ+b sin⁡θ )^2=c^2a^2  〖cos〗^2⁡θ+b^2  〖sin〗^2⁡θ+2ab sin⁡θ  cos⁡θ=c^2….(1) Now let a sin⁡θ-b cos⁡θ=ksquaring both sides, we geta^2  〖sin〗^2⁡θ+b^2  〖cos〗^2⁡θ-2ab sin⁡θ  cos⁡θ=k^2….(2) Adding (1)and (2)a^2 (〖cos〗^2⁡θ+〖sin〗^2⁡θ )+b^2 (〖cos〗^2⁡θ+〖sin〗^2⁡θ )=c^2+k^2a^2+b^2=c^2+k^2k=±(a^2+b^2-c^2 )^(1/2)