Question

if cos theta + sin theta =√2 cos theta then prove that cos that - sin theta = √2 sin theta ​

Answer 1

$\huge{\underline{\mathcal{\red{A}\green{n}\pink{s}\orange{w}\blue{e}\pink{R:-}}}}$

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$\large\underline\bold\red{Given:-}$

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

$\large\underline\bold\red{To\:Prove:-}$

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

$\large\underline\bold\red{Solution:-}$

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

Squaring on both sides we get,

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

$\implies { { \cos }^{2} \theta } + { \sin }^{2} \theta + 2 \sin \theta \cos\theta = 2 { \cos }^{2} \theta$

$\implies 2 { \cos }^{2} \theta - {cos}^{2} \theta - {sin}^{2} \theta = 2sin \theta cos \theta$

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

$\implies (cos \theta + sin \theta)(cos \theta - sin \theta) = 2sin \theta \: cos \theta$

We know that ,

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

Therefore,

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

$\implies (cos \theta - sin \theta) = 2sin \theta \: cos \theta \div \sqrt{2}cos \theta$

${\boxed{\bold{\small{\implies cos \theta - sin \theta = \sqrt{2}sin \theta}}}}$

Hence, Proved

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$\large\underline\bold\red{Trignometric\:identities:-}$

ㅤㅤㅤ$\boxed{\begin{minipage}{4cm}\\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

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Answer 2

Answer:

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Explanation:

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