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If $\sin A + \sin^2 A = 1$, find the value of $\cos^2 A + \cos^4 A$. Also, using the above, prove that $\tan^2 A \cdot \sec^2 A = 1$.
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Solution: (a) $\sin A + \sin^2 A = 1$
$\implies \sin A = \cos^2 A$ (1/2 Mark)
$\sin^2 A = \cos^4 A$ (On squaring both sides) (1/2 Mark)
$1 - \cos^2 A = \cos^4 A \implies \cos^2 A + \cos^4 A = 1$ (1 Mark)
LHS $= \tan^2 A \cdot \sec^2 A = \frac{\sin^2 A}{\cos^2 A} \times \frac{1}{\cos^2 A} = \frac{\sin^2 A}{\sin^2 A} \times 1 = 1 = \text{RHS}$ ($:: \cos^4 A = \sin^2 A$) (1 Mark)
$\implies \sin A = \cos^2 A$ (1/2 Mark)
$\sin^2 A = \cos^4 A$ (On squaring both sides) (1/2 Mark)
$1 - \cos^2 A = \cos^4 A \implies \cos^2 A + \cos^4 A = 1$ (1 Mark)
LHS $= \tan^2 A \cdot \sec^2 A = \frac{\sin^2 A}{\cos^2 A} \times \frac{1}{\cos^2 A} = \frac{\sin^2 A}{\sin^2 A} \times 1 = 1 = \text{RHS}$ ($:: \cos^4 A = \sin^2 A$) (1 Mark)