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Question Number 86490 by ram roop sharma last updated on 29/Mar/20

$$\mathrm{If}\sin \theta +\cos \theta =\sqrt{2}\cos \theta \mathrm{then}$$$$\cos \theta -\sin \theta \mathrm{is}\mathrm{equal}\mathrm{to}$$

Commented by john santu last updated on 29/Mar/20

$$\sin \theta =\sqrt{2}\cos \theta -\cos \theta$$$$\tan \theta =\sqrt{2}-1$$$$\mathrm{let}\cos \theta -\sin \theta =\mathrm{k}$$$$\sin \theta -\cos \theta =-\mathrm{k}$$$$(\sin \theta -\cos \theta )(\sin \theta +\cos \theta )=-\mathrm{k}\sqrt{2}\cos \theta$$$$-\cos 2\theta =-\mathrm{k}\sqrt{2}\cos \theta$$$$2\cos {}^2\theta -1=\mathrm{k}\sqrt{2}\cos \theta$$$$\mathrm{k}=\frac{2\cos {}^2\theta -1}{\sqrt{2}\cos \theta }=\sqrt{2}\cos \theta -\frac{1}{\sqrt{2}}\cos \theta$$$$=\sqrt{2}\left(\frac{1}{\sqrt{2}\left(\sqrt{2-\sqrt{2}}\right)}\right)-\frac{1}{2\sqrt{2-\sqrt{2}}}$$$$=\frac{1}{\sqrt{2-\sqrt{2}}}-\frac{1}{2\sqrt{2-\sqrt{2}}}=\frac{1}{2\sqrt{2-\sqrt{2}}}$$

Answered by som(math1967) last updated on 29/Mar/20

$$sin\theta =(\sqrt{2}-1)cos\theta$$$$(\sqrt{2}+1)sin\theta =(\sqrt{2}+1)(\sqrt{2}-1)cos\theta$$$$\sqrt{2}sin\theta +sin\theta =1.cos\theta$$$$\sqrt{2}sin\theta =cos\theta -sin\theta$$$$\therefore cos\theta -sin\theta =\sqrt{2}sin\theta ans$$

Commented by peter frank last updated on 29/Mar/20

$$thankyou$$

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