Question

If cos theta - tan theta = root 2 sin theta then prove that cos theta + sin theta = 2 root cos theta

Answer 1

Step-by-step explanation:

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$\huge{\underline{\mathrm{Question}}}$

If cos theta - tan theta = root 2 sin theta then prove that cos theta + sin theta = 2 root cos theta

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$\huge{\underline{\mathrm{Answer}}}$

$\cos(0) + \sin(0) = \sqrt{2} \cos(0)$

∴ squaring on the both sides, We get⇒

∴⇛$\cos {}^{2} 0 + \sin {}^{2} 0 +2 \sin0 \cos0 {}^{} = 2 \cos {}^{2} 0$

∴$⇛\cos {}^{2} 0 - \sin {}^{2} 0 = 2 \cos0 \: \sin0$

∴⇛$( \cos 0 + \sin 0 = \cos 0 - \sin 0 ) = 2 \cos0 \sin0$

∴⇛$\sqrt{2} \cos \: ( \cos0 - \sin0) = 2 \cos0 \sin0$

∴$[Given \cos0 + \sin0 = \sqrt{2} \cos0]$

$\huge{\underline{\mathrm{Hence\: Proved}}}$

∴$\sf\blue{ \cos0 - \sin0 = \sqrt{2} \sin0}$

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$\huge{\underline{\mathrm{Note}}}$

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Answer 2

Question

If cosθ -sinθ=√2sinθ

then prove that cos θ +sin θ =√2 cos θ

$\sf\huge\bold{Solution}$

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

Squaring both sides:

(cosθ -sinθ)²=(√2sinθ)²

$\sf \longmapsto \: {cos}^{2} \theta + {sin}^{2} \theta - 2sin \theta \: cos \theta \: = 2 {sin}^{2} \theta$

$\sf \longmapsto1 - 2sin \theta \: cos \theta = 2 {sin}^{2} \theta$

$\sf \longmapsto2sin \theta \: cos \theta = 1 - 2 {sin}^{2} \theta$

Taking L.H.S

$\sf \longmapsto {(sin \theta + cos \theta)}^{2} = {sin}^{2} \theta + {cos}^{2} \theta + 2sin \theta \: cos \theta$

$\sf \longmapsto1 + 1 - 2 {sin}^{2} \theta$

$\sf \longmapsto2(1 - {sin}^{2} \theta) = 2 {cos}^{2} \theta$

$\sf{(sin \theta + cos \theta)}^{2} = 2 {cos}^{2} \theta$

$\sf \: sin \theta + cos \theta = \sqrt{2} cos \theta(RHS)$