If tan theta = frac{sin alpha - cos alpha}{si | vedclass

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(A) Given $	an 	heta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$.Dividing numerator and denominator by $\cos \alpha$,we get $	an

Vedclass · Accepted Answer

$\sqrt{2} \cos 	heta, \sqrt{2} \sin 	heta$

Vedclass · Answer

(A) Given $	an 	heta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$.Dividing numerator and denominator by $\cos \alpha$,we get $	an 	heta = \frac{	an \alpha - 1}{	an \alpha + 1} = 	an(\alpha - \frac{\pi}{4})$.Thus,$	heta = \alpha - \frac{\pi}{4}$,which implies $\alpha = 	heta + \frac{\pi}{4}$.Now,$\sin \alpha + \cos \alpha = \sin(	heta + \frac{\pi}{4}) + \cos(	heta + \frac{\pi}{4}) = (\sin 	heta \cos \frac{\pi}{4} + \cos 	heta \sin \frac{\pi}{4}) + (\cos 	heta \cos \frac{\pi}{4} - \sin 	heta \sin \frac{\pi}{4}) = \frac{1}{\sqrt{2}}(\sin 	heta + \cos 	heta + \cos 	heta - \sin 	heta) = \frac{2 \cos 	heta}{\sqrt{2}} = \sqrt{2} \cos 	heta$.Similarly,$\sin \alpha - \cos \alpha = \sin(	heta + \frac{\pi}{4}) - \cos(	heta + \frac{\pi}{4}) = (\sin 	heta \cos \frac{\pi}{4} + \cos 	heta \sin \frac{\pi}{4}) - (\cos 	heta \cos \frac{\pi}{4} - \sin 	heta \sin \frac{\pi}{4}) = \frac{1}{\sqrt{2}}(\sin 	heta + \cos 	heta - \cos 	heta + \sin 	heta) = \frac{2 \sin 	heta}{\sqrt{2}} = \sqrt{2} \sin 	heta$.

If $\tan \theta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$,then $\sin \alpha + \cos \alpha$ and $\sin \alpha - \cos \alpha$ are equal to:

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