13. If sin θ+ cos θ = √2 then θ=
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[tex]\huge\mathcal\colorbox{lavender}{{\color{b}{✿Yøur-Añswer♡}}}[/tex]
[tex]\large\bf{\underline{\red{Maths}}}[/tex]
The answer is 45°
For explaining refer to the above attachment
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Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\rm \:sin\theta + cos\theta = \sqrt{2} \\ [/tex]
can be rewritten as
[tex]\rm \: cos\theta = \sqrt{2} - sin\theta \\ [/tex]
On squaring both sides, we get
[tex]\rm \: (cos\theta)^{2} = (\sqrt{2} - sin\theta)^{2} \\ [/tex]
[tex]\rm \: {cos}^{2}\theta = {( \sqrt{2}) }^{2} + {sin}^{2}\theta - 2 \times \sqrt{2} \times sin\theta \\ [/tex]
[tex]\rm \: {cos}^{2}\theta = 2 + {sin}^{2}\theta - 2 \sqrt{2}sin\theta \\ [/tex]
can be further rewritten as
[tex]\rm \: 1 - {sin}^{2}\theta = 2 + {sin}^{2}\theta - 2 \sqrt{2}sin\theta \\ [/tex]
[tex]\rm \: 2{sin}^{2}\theta - 2 \sqrt{2}sin\theta + 1 = 0 \\ [/tex]
can be rewritten as
[tex]\rm \: {( \sqrt{2} sin\theta )}^{2} - 2 \times \sqrt{2}sin\theta \times 1 + {(1)}^{2} = 0 \\ [/tex]
[tex]\rm \: {( \sqrt{2}sin\theta - 1) }^{2} = 0 \\ [/tex]
[tex]\rm \: \sqrt{2} sin\theta - 1 = 0 \\ [/tex]
[tex]\rm \: \sqrt{2} sin\theta = 1\\ [/tex]
[tex]\rm \: sin\theta = \dfrac{1}{ \sqrt{2} } \\ [/tex]
[tex]\rm \: sin\theta =sin45\degree \\ [/tex]
[tex]\rm\implies \:\theta = 45\degree \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information :-
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \\ \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]