Question

If cos theta - sin theta = (root 2)(sin theta) then prove that cos theta + sin theta = (root 2)(cos theta).

Answer 1

Theta is written as A .


Given, cosA - sinA = √2 sinA

Divide by sinA on both sides.

( cosA - sinA ) ÷ sinA = √2 sinA ÷ sinA

cosA / sinA - sinA / sinA = √2

cot A - 1 = √2

cotA = √2 + 1

1 / tanA = √2 + 1

tanA = 1 / { √2 + 1 }

tanA = 1 / ( √2 + 1 ) × ( √2 - 1 ) / ( √2 - 1 )

tanA = ( √2 - 1 ) / ( 2 - 1 )

tanA = √2 - 1

tan A + 1 = √2

sinA / cosA + 1 = √2

( sinA + cosA ) / cosA = √2

sinA + cosA = √2 cosA

Answer 2

$\underline{\large\bf{\mathfrak{Hello!}}}$

$Given \: that: \\\\ => cos\theta - sin\theta = \sqrt{2} sin\theta\\\\\\Squaring\:both\:sides,\:we\:get\\\\ => {(cos\theta - sin\theta)}^{2} = {(\sqrt{2} sin\theta)}^{2}\\\\ =>{cos}^{2}\theta +{sin}^{2}\theta - 2sin\theta cos\theta = 2{sin}^{2}\theta \\\\ => {cos}^{2}\theta +{sin}^{2}\theta -2{sin}^{2}\theta = 2sin\theta cos\theta\\\\ => {cos}^{2}\theta -{sin}^{2}\theta = 2sin\theta cos\theta\\\\ => (cos\theta + sin\theta)(cos\theta - sin\theta) = 2sin\theta cos\theta\\\\ => (cos\theta + sin\theta)(\sqrt{2} sin\theta) = 2sin\theta cos\theta \\\\ => cos\theta + sin\theta = \frac{2sin\theta cos\theta}{\sqrt{2} sin\theta} \\\\ => cos\theta + sin\theta = \sqrt{2} cos\theta \\\\ Hence \: Proved$

$<marquee> Hope this helps...:)$