If \(a\, \cos \theta+b\, \sin \, \theta=2\) and \(a\, \sin \, \theta-b\, \cos \, \theta=3\) , then \({a}^{2^{}}+{b}^2=\)
Square and add both the equations: \[ a^2 (\sin^2 \theta + \cos^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = 2^2 + 3^2 \] Since \( \sin^2 \theta + \cos^2 \theta = 1 \), the equation simplifies to: \[ a^2 + b^2 = 13 \]
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