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 - 45^\circ)$.Thus,$	heta = \alpha - 45^\circ$,which implies $\alpha = 	heta + 45^\circ$.Now,$\sin \alpha + \cos \alpha = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \alpha + \frac{1}{\sqrt{2}} \cos \alpha ight) = \sqrt{2} \sin(\alpha + 45^\circ) = \sqrt{2} \sin(	heta + 45^\circ + 45^\circ) = \sqrt{2} \sin(	heta + 90^\circ) = \sqrt{2} \cos 	heta$.Similarly,$\sin \alpha - \cos \alpha = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \alpha - \frac{1}{\sqrt{2}} \cos \alpha ight) = \sqrt{2} \sin(\alpha - 45^\circ) = \sqrt{2} \sin(	heta + 45^\circ - 45^\circ) = \sqrt{2} \sin 	heta$.

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

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