frac{sqrt{2}-sin alpha-cos alpha}{sin alpha-c | vedclass

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Question

(C) Given expression is $E = \frac{\sqrt{2}-(\sin \alpha+\cos \alpha)}{\sin \alpha-\cos \alpha}$.Multiply numerator and denominator by $\frac{1}{\sqrt

Vedclass · Accepted Answer

$	an \left(\frac{\alpha}{2}-\frac{\pi}{8}ight)$

Vedclass · Answer

(C) Given expression is $E = \frac{\sqrt{2}-(\sin \alpha+\cos \alpha)}{\sin \alpha-\cos \alpha}$.Multiply numerator and denominator by $\frac{1}{\sqrt{2}}$:$E = \frac{1-\frac{1}{\sqrt{2}}(\sin \alpha+\cos \alpha)}{\frac{1}{\sqrt{2}}(\sin \alpha-\cos \alpha)}$.Using $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get:$E = \frac{1-(\sin \alpha \cos \frac{\pi}{4} + \cos \alpha \sin \frac{\pi}{4})}{\sin \alpha \cos \frac{\pi}{4} - \cos \alpha \sin \frac{\pi}{4}} = \frac{1-\sin(\alpha+\frac{\pi}{4})}{\sin(\alpha-\frac{\pi}{4})}$.Using $1-\sin 	heta = 1-\cos(\frac{\pi}{2}-	heta) = 2\sin^2(\frac{\pi}{4}-\frac{	heta}{2})$ and $\sin 	heta = 2\sin(\frac{	heta}{2})\cos(\frac{	heta}{2})$:$E = \frac{2\sin^2(\frac{\pi}{4}-\frac{\alpha}{2}-\frac{\pi}{8})}{2\sin(\frac{\alpha}{2}-\frac{\pi}{8})\cos(\frac{\alpha}{2}-\frac{\pi}{8})} = \frac{\sin^2(\frac{\pi}{8}-\frac{\alpha}{2})}{\sin(\frac{\alpha}{2}-\frac{\pi}{8})\cos(\frac{\alpha}{2}-\frac{\pi}{8})}$.Since $\sin^2(\frac{\pi}{8}-\frac{\alpha}{2}) = \sin^2(\frac{\alpha}{2}-\frac{\pi}{8})$:$E = \frac{\sin(\frac{\alpha}{2}-\frac{\pi}{8})}{\cos(\frac{\alpha}{2}-\frac{\pi}{8})} = 	an(\frac{\alpha}{2}-\frac{\pi}{8})$.

$\frac{\sqrt{2}-\sin \alpha-\cos \alpha}{\sin \alpha-\cos \alpha}=$

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