If \(2\sin^{2}A+2\sqrt{2}\cos A-3=0\), then find the value of tanA.
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The equation carries both \(\sin^{2}A\) and \(\cos A\). To solve it we must express everything through a single ratio, so replace the sine-square using the Pythagorean identity \(\sin^{2}A=1-\cos^{2}A\).
Substituting into \(2\sin^{2}A+2\sqrt{2}\cos A-3=0\) gives \(2(1-\cos^{2}A)+2\sqrt{2}\cos A-3=0\).
Expand the bracket: \(2-2\cos^{2}A+2\sqrt{2}\cos A-3=0\), which tidies to \(-2\cos^{2}A+2\sqrt{2}\cos A-1=0\).
Multiply by \(-1\) to make the leading coefficient positive: \(2\cos^{2}A-2\sqrt{2}\cos A+1=0\), a quadratic in \(\cos A\).
Its discriminant is \(D=(2\sqrt{2})^{2}-4\cdot2\cdot1=8-8=0\), signalling one repeated root \(\cos A=\dfrac{2\sqrt{2}}{2\cdot2}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\).
The angle in the first quadrant with \(\cos A=\dfrac{1}{\sqrt{2}}\) is \(A=45^{\circ}\), since \(\cos 45^{\circ}=\dfrac{1}{\sqrt{2}}\).
Therefore \(\tan A=\tan 45^{\circ}=\dfrac{\sin 45^{\circ}}{\cos 45^{\circ}}=\dfrac{1/\sqrt{2}}{1/\sqrt{2}}=1\).
The concept is using \(\sin^{2}A=1-\cos^{2}A\) to convert to a solvable quadratic in cosine, then evaluating the required ratio at the standard angle.
Hence the value of tan A is 1.
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