Question

If CosA+SinA=1, Prove that CosA-SinA=+_1

Answer 1

$\cos(a) + \sin(a) = 1 \\ squaring \: both \: sides \\ { \cos }^{2} a + {sin}^{2} a + 2sinacosa = 1 \\ 1 - {sin}^{2} a + 1 - {cos}^{2} a + 2sinacosa = 1\: (using \: {sin}^{2} + {cos}^{2} = 1 ) \\ 2 - {sin}^{2} a - {cos}^{2} a + 2cosasina = 1 \\ {sin}^{2} a + {cos }^{2} - 2cosasina = 1 \\ ({cosa - sina})^{2} = 1 \\ cosa - sina = + - 1......hence \: proved$

Answer 2

Step-by-step explanation:

If

$cos \: A + sin \: A = 1 ...eq1$

To prove

$cos \: A - sin \: A = ± 1$

squaring both side eq1

$( {cos \: A + sin \: A)}^{2} = ( {1)}^{2} \\ \\ {cos}^{2} A + {sin}^{2} A + 2sin \: A \: cos \: A = 1 \\ \\ as \\ \\ {cos}^{2} A=1- {sin}^{2} A \\\\ {sin}^{2} A=1- {cos}^{2} A \\ \\ 1 - {sin}^{2} A + 1 - {cos}^{2} A + 2sin \: A \: cos \: A = 1 \\ \\ 2 - {sin}^{2} A - {cos}^{2} A + 2sin \: A \: cos \: A = 1 \\ \\ - {sin}^{2} A - {cos}^{2} A + 2sin \: A \: cos \: A = 1 - 2 \\ \\ {sin}^{2} A + {cos}^{2} A - 2sin \: A \: cos \: A = 1 \\ \\ {(cos \: A - sin \: A)}^{2} = 1 \\ \\ taking \: square \: root \: both \: sides \\ \\ cos \: A - sin \: A = ±1$

Hence proved.

Hope it helps you.