If a cos $\theta$ + b sin $\theta$ = 4 and a sin $\theta$ − b sin $\theta$ = 3, then a2 + b2 =

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KollegeApply · Accepted Answer

Given: a cos $\theta$+b sin$\theta$=4 a sin $\theta$-b cos$\theta$=3 Squaring and then adding the above two equations, we have `(a cos$\theta$+b sin$\theta$)^2+(a sin$\theta$-b cos$\theta$)^2=(4)^2+(3)^2 =(a^2cos^2$\theta$+b^2 sin^2$\theta$+2a cos$\theta$.b.sin$\theta$)+(a^2 sin^2$\theta$+b^2 cos^2$\theta$-2.a sin$\theta$.b cos$\theta$)=16+9 =a^2 cos^2$\theta$+b^2 sin^2$\theta$+ab sin$\theta$ cos$\theta$+a^2 sin^2$\theta$+b^2 cos^2$\theta$-2ab sin$\theta$ cos$\theta$=25 =a^2 cos^2$\theta$+b^2 sin^2$\theta$+a^2 sin^2$\theta$+b^2 cos^2$\theta$=25 =(a^2 cos^2$\theta$+a^2sin^2$\theta$)+(b^2 sin^2$\theta$+b^2 cos^2$\theta$)=25 =a^2(cos^2$\theta$+sin^2$\theta$)+b^2(sin^2$\theta$+cos^2$\theta$=25) =a^2(1)+b^2(1)=25 =a^2+b^2=25``

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