Question

if a cos theta + b sin theta equal to 4 and a sin theta minus b cos theta equal to 3 find a square + b square​

Answer 1

$a cos\theta+bsin\theta=4\\ \\ \\(a cos\theta+bsin\theta)^{2}=16\\ \\ \\a^2cos^2\theta+b^2sin^2\theta+2abcos\theta sin\theta=16\\ \\ \\ \\asin\theta-bcos\theta=3\\ \\ \\(asin\theta-bcos\theta)^2=9\\ \\ \\a^2sim^2\theta+b^2cos^2\theta-2absin\theta cos\theta=9\\ \\ \\ \\ \\ \\Adding\:two\:equations:-\\ \\ \\a^2cos^2\theta+b^2sin^2\theta+2abcos\theta sin\theta\\+a^2sin^2\theta+b^2cos^2\theta-2absin\theta cos\theta=25\\ \\ \\(a^2cos^2\theta+b^2cos^2\theta)+(a^2sin^2\theta+b^2sin^2\theta)=25$

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$cos^2\theta(a^2+b^2)+sin^2\theta(a^2+b^2)=25\\ \\ \\(a^2+b^2)(sin^2\theta+cos^2\theta)=25\\ \\ \\a^2+b^2=25\\ \\(Since,sin^2x+cos^2x=1)$

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