Show that ${(1 - sin A + cos A)}^{2} = 2 (1 - sin A) (1 + cos A)$ .

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Solution

Considering LHS:
${(1 - sin A + cos A)}^{2} = 1 + {sin}^{2} A + {cos}^{2} A - 2 sin A + 2 cos A - 2 sin A cos A$
$= 2 - 2 sin A + 2 cos A - 2 sin A cos A$

Considering RHS:
$2 (1 - sin A) (1 + cos A) = 2 (1 + cos A - sin A - sin A cos A)$
$= 2 + 2 cos A - 2 sin A - 2 sin A cos A$
$2 - 2 sin A + 2 cos A - 2 sin A cos A = 2 + 2 cos A - 2 sin A - 2 sin A cos A$
$∴ L H S = R H S$

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