Question

Class 10th maths . question no. 18 please.

Answer 1

$\bold{\large{\underline{\underline{ Step - by - step \:Solution\: \colon}}}}$

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$\bold{\large{Given \: \colon \: cos \: \theta \: - \: sin\: \theta = \sqrt{2} sin\: \theta }}$

$\bold{\large{To \: Prove\: \colon \: cos \: \theta \: + \: sin\: \theta \: = \: \sqrt{2} cos\: \theta }}$

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$\bold{\large{\underline{Solution\: \colon}}}$

Squaring both sides :

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

$\bold{\large{ cos^{2} \: \theta + sin^{2} \: \theta - 2 sin\: \theta cos\: \theta = 2 sin^{2} \: \theta }}$

$\bold{\large{ cos^{2} \: \theta + sin^{2}\: \theta - 2 sin^{2}\: \theta = 2 sin\: \theta \: cos \: \theta }}$

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

$\bold{\large{ (cos\: \theta + sin\: \theta)(cos\: \theta - sin\: \theta )= 2 sin\: \theta \: cos \: \theta }}$

.......{eqt 1}

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$\bold{\large{\therefore\: cos \:\theta - sin\: \theta = \sqrt{2}\: sin \theta}}$

Put the given value in equation :

$\bold{\large{ (\sqrt{2}sin\: \theta)(cos\: \theta + sin\: \theta) = 2 sin\: \theta \:cos \: \theta }}$

$\bold{\large{ cos\: \theta + sin \: \theta = \dfrac{2\:sin\: \theta \:cos \: \theta}{\sqrt{2} sin\: \theta}}}$

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

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As according to this solution becomes :

$\bold{\large{Answer}}$

$\bold{\boxed{\large{ cos \: \theta + sin\: \theta = \sqrt{2} cos \: \theta}}}$

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Hence proved

[Assumption error in the question]

$\bold{\large{Thanks}}$

Answer 2

It seems that the given question is incorrect.


Correct question : Prove that cosA + sinA = √2 cosA



$\mathsf{solution :}$


cosA - sinA = √2 sinA

$\dfrac{ \cos A - \sin A }{ \sin A} = \sqrt{2} \\ \\ \\ \dfrac{ \cos A }{ \sin A} - \frac{ \sin A }{ \sin A } = \sqrt{2} \\ \\ \\ \cot \: A \: - 1 = \sqrt{2} \\ \\ \cot \: A= \sqrt{2} + 1 \\ \\ \\ Therefore, \\ \\ \\ \dfrac{1}{ \cot A} = \frac{1}{ \sqrt{2} + 1 } \\ \\ \\ \frac{1}{\cot A} = \sqrt{ 2 }- 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{by \: rationalization} \\ \\ \\ \tan A = \sqrt{2} - 1 \\ \\ tanA + 1 = \sqrt{2}$



Multiply by cosA on both sides,



$(\dfrac{ \sin A }{ \cos A} \times \cos \: A) + (1 \times \cos A) = \sqrt{2} \cos A$


sinA + cosA = √2 cosA



Proved.