if tan theta=(a/b) then prove that (a sin theta+b cos theta)=[(a^2)-(b^2)]/[(a^2)+(b^2)]

Question

math trigonometry problem pls help!!!

Answer 1

Sin(theta)=a/sqrt(a^2+b^2); cos(theta)=b/sqrt(a^2+b^2).

asin(theta)=a^2/sqrt(a^2+b^2); bcos(theta)=b^2/sqrt(a^2+b^2).

asin(theta)+bcos(theta)=(a^2+b^2)/sqrt(a^2+b^2)=sqrt(a^2+b^2).

Consider a=b, then this would be sqrt(2a^2)=asqrt(2), whereas the expression in the question would evaluate as zero, disproving the supposed equality. Also, if a=0 but b is not equal to zero the expression would evaluate to -1 whereas sqrt(a^2+b^2)=b (sin(theta)=0 and cos(theta)=1, so asin(theta)+bcos(theta)=b).

(a^2-b^2)/(a^2+b^2)=a^2/(a^2+b^2)-b^2/(a^2+b^2)=sin^2(theta)-cos^2(theta)=

(sin(theta)-cos(theta))(sin(theta)+cos(theta)).