If cos theta + sin theta = sqrt{2} cos theta | vedclass

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(A) Given $\cos 	heta + \sin 	heta = \sqrt{2} \cos 	heta$. Dividing by $\cos 	heta$,we get $1 + 	an 	heta = \sqrt{2}$. So,$	an 	heta = \sqrt{2

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$\cot 	heta$

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(A) Given $\cos 	heta + \sin 	heta = \sqrt{2} \cos 	heta$. Dividing by $\cos 	heta$,we get $1 + 	an 	heta = \sqrt{2}$. So,$	an 	heta = \sqrt{2} - 1$. We need to find $\sec 2 	heta + 	an 2 	heta = \frac{1}{\cos 2 	heta} + \frac{\sin 2 	heta}{\cos 2 	heta} = \frac{1 + \sin 2 	heta}{\cos 2 	heta}$. Using the identity $	an 	heta = \frac{\sin 2 	heta}{1 + \cos 2 	heta}$,we have $\frac{1 + \sin 2 	heta}{\cos 2 	heta} = \frac{(\cos 	heta + \sin 	heta)^2}{\cos^2 	heta - \sin^2 	heta} = \frac{(\cos 	heta + \sin 	heta)^2}{(\cos 	heta - \sin 	heta)(\cos 	heta + \sin 	heta)} = \frac{\cos 	heta + \sin 	heta}{\cos 	heta - \sin 	heta}$. Dividing numerator and denominator by $\cos 	heta$,we get $\frac{1 + 	an 	heta}{1 - 	an 	heta}$. Substituting $	an 	heta = \sqrt{2} - 1$,we get $\frac{1 + (\sqrt{2} - 1)}{1 - (\sqrt{2} - 1)} = \frac{\sqrt{2}}{2 - \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}(\sqrt{2} - 1)} = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$. Since $	an 	heta = \sqrt{2} - 1$,then $\cot 	heta = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$. Thus,$\sec 2 	heta + 	an 2 	heta = \cot 	heta$.

If $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$ and $0 < \theta < \frac{\pi}{2}$,then $\sec 2 \theta + \tan 2 \theta = $

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