If \(\sin A = \dfrac{3}{5}\) and A lies in the second quadrant, find the value of \((\sin A + \cos A)^2\).
\(\dfrac{1}{25}\)
Given \(\sin A = \dfrac{3}{5}\), A in Q2.
In Q2, cosine is negative: \(\cos A = -\sqrt{1-\sin^2 A} = -\dfrac{4}{5}\)
\((\sin A + \cos A)^2 = \left(\dfrac{3}{5} - \dfrac{4}{5}\right)^2 = \left(-\dfrac{1}{5}\right)^2 = \mathbf{\dfrac{1}{25}}\)
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