Question

13. If sin θ+ cos θ = √2 then θ=


tq​

Answer 1

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$\large\bf{\underline{\red{Maths}}}$

The answer is 45°

For explaining refer to the above attachment

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \:sin\theta + cos\theta = \sqrt{2}$

can be rewritten as

$\rm \: cos\theta = \sqrt{2} - sin\theta$

On squaring both sides, we get

$\rm \: (cos\theta)^{2} = (\sqrt{2} - sin\theta)^{2}$

$\rm \: {cos}^{2}\theta = {( \sqrt{2}) }^{2} + {sin}^{2}\theta - 2 \times \sqrt{2} \times sin\theta$

$\rm \: {cos}^{2}\theta = 2 + {sin}^{2}\theta - 2 \sqrt{2}sin\theta$

can be further rewritten as

$\rm \: 1 - {sin}^{2}\theta = 2 + {sin}^{2}\theta - 2 \sqrt{2}sin\theta$

$\rm \: 2{sin}^{2}\theta - 2 \sqrt{2}sin\theta + 1 = 0$

can be rewritten as

$\rm \: {( \sqrt{2} sin\theta )}^{2} - 2 \times \sqrt{2}sin\theta \times 1 + {(1)}^{2} = 0$

$\rm \: {( \sqrt{2}sin\theta - 1) }^{2} = 0$

$\rm \: \sqrt{2} sin\theta - 1 = 0$

$\rm \: \sqrt{2} sin\theta = 1$

$\rm \: sin\theta = \dfrac{1}{ \sqrt{2} }$

$\rm \: sin\theta =sin45\degree$

$\rm\implies \:\theta = 45\degree$

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Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \\ \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$