frac{1+sin theta-cos theta}{1+sin theta+cos t | vedclass

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(B) We use the identities: $1 - \cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$,$1 + \cos 	heta = 2 \cos^2 \left(\frac{	heta}{2}ight)$,and $

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$	an \left(\frac{	heta}{2}ight)$

Vedclass · Answer

(B) We use the identities: $1 - \cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$,$1 + \cos 	heta = 2 \cos^2 \left(\frac{	heta}{2}ight)$,and $\sin 	heta = 2 \sin \left(\frac{	heta}{2}ight) \cos \left(\frac{	heta}{2}ight)$.Substituting these into the expression:$= \frac{2 \sin^2 \left(\frac{	heta}{2}ight) + 2 \sin \left(\frac{	heta}{2}ight) \cos \left(\frac{	heta}{2}ight)}{2 \cos^2 \left(\frac{	heta}{2}ight) + 2 \sin \left(\frac{	heta}{2}ight) \cos \left(\frac{	heta}{2}ight)}$$= \frac{2 \sin \left(\frac{	heta}{2}ight) [\sin \left(\frac{	heta}{2}ight) + \cos \left(\frac{	heta}{2}ight)]}{2 \cos \left(\frac{	heta}{2}ight) [\cos \left(\frac{	heta}{2}ight) + \sin \left(\frac{	heta}{2}ight)]}$$= \frac{\sin \left(\frac{	heta}{2}ight)}{\cos \left(\frac{	heta}{2}ight)} = 	an \left(\frac{	heta}{2}ight)$.

$\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}$ is equal to

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