Question

if cosθ+sinθ=√2,then cosθ-sinθ=

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Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ cos \: x - sin \: x \\ \\ given \: that \\ cos \: x + sin \: x = \sqrt{2} \\ cos \: x = \sqrt{2} - sin \: x \: \: \: \: \: (1) \\ \\ we \: know \: that \\ sin {}^{2} + cos {}^{2} x = 1 \\ sin {}^{2} \: x + ( \sqrt{2} - sin \: x) {}^{2} = 1 \\ \\ sin {}^{2} \: x + sin {}^{2} \: x + 2 - 2 \sqrt{2} .sin \: x = 1 \\ \\ 2.sin {}^{2} \: x - \sqrt{2} .sin \: x + 1 = 0 \\ \\ 2.sin {}^{2} \: x - \sqrt{2} .sin \: x - \sqrt{2} .sin \: x + 1 = 0 \\ \\ \sqrt{2} .sin \: x( \sqrt{2} .sin \: x - 1) - 1( \sqrt{2} .sin \: x - 1) = 0 \\ \\ ( \sqrt{2} .sin \: x - 1)( \sqrt{2} .sin \: x - 1) = 0 \\ \\ \sqrt{2} .sin \: x - 1 = 0 \\ \\ sin \: x = \frac{1}{ \sqrt{2} }$

$thus \: then \: we \: know \: that \\ si n \: 45 = \frac{1}{ \sqrt{2} } \\ \\ thus \: then \: accordingly \\ cos \: 45 = \frac{1}{ \sqrt{2} } \\ \\ so \: hence \\ cos \: x - sin \: x = \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{2} } \\ \\ cos \: x - sin \: x = 0$