Prove that $\sec A(1-\sin A)(\sec A+\tan A)=1$.
(A) $LHS = \sec A(1-\sin A)(\sec A+\tan A)$
$= \left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$
$= \left(\frac{1-\sin A}{\cos A}\right)\left(\frac{1+\sin A}{\cos A}\right)$
$= \frac{(1-\sin A)(1+\sin A)}{\cos^2 A}$
$= \frac{1-\sin^2 A}{\cos^2 A}$
$= \frac{\cos^2 A}{\cos^2 A} = 1 = RHS$
Hence,the identity is proved.
$= \left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$
$= \left(\frac{1-\sin A}{\cos A}\right)\left(\frac{1+\sin A}{\cos A}\right)$
$= \frac{(1-\sin A)(1+\sin A)}{\cos^2 A}$
$= \frac{1-\sin^2 A}{\cos^2 A}$
$= \frac{\cos^2 A}{\cos^2 A} = 1 = RHS$
Hence,the identity is proved.
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