Question

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Answer 1

Answer:

$sin \theta + cos \theta = \sqrt{2} \\ \therefore \: sin \theta = \sqrt{2} - cos \theta \\ squaring \: both \: sides \\ {sin}^{2} \theta = {(\sqrt{2} - cos \theta )}^{2} \\ \therefore \: {sin}^{2} \theta =2 + {cos}^{2} \theta - 2 \sqrt{2} cos \theta \\ \therefore \: 1 - {cos}^{2} \theta =2 + {cos}^{2} \theta - 2 \sqrt{2} cos \theta \\ \therefore \:0 = 2 + {cos}^{2} \theta - 2 \sqrt{2} cos \theta - 1 + {cos}^{2} \theta \\ \therefore \:0 = 2{cos}^{2} \theta - 2 \sqrt{2} cos \theta + 1 \\ or \\ \: \: \: \: \: 2{cos}^{2} \theta - 2 \sqrt{2} cos \theta + 1 = 0 \\ \therefore \: 2{cos}^{2} \theta - \sqrt{2} cos \theta - \sqrt{2} cos \theta+ 1 = 0 \\ \sqrt{2} cos \theta( \sqrt{2} cos \theta - 1) - 1( \sqrt{2} cos \theta - 1) = 0 \\ ( \sqrt{2} cos \theta - 1) ( \sqrt{2} cos \theta - 1)= 0 \\ {( \sqrt{2} cos \theta - 1)}^{2} = 0 \\ taking \: square \: root \: on \: both \: sides \\ \sqrt{2} cos \theta - 1 = 0 \\ \implies \: \sqrt{2} cos \theta = 1 \\ \implies \: cos \theta = \frac{1}{\sqrt{2} } \\ \implies \: cos \theta = cos \: 45 \degree \\ \implies \huge \fbox{ \theta = 45 \degree } \\ \implies \huge \fbox{tan \theta = tan \: 45 \degree = 1}$