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

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Question

(B) આપણે નિત્યસમ $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ અને $\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1 = 1 - 2 \sin^2 \frac{\

Vedclass · Accepted Answer

$-\cot \frac{	heta}{2}$

Vedclass · Answer

(B) આપણે નિત્યસમ $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ અને $\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1 = 1 - 2 \sin^2 \frac{	heta}{2}$ નો ઉપયોગ કરીએ છીએ.આ કિંમતો પદાવલિમાં મૂકતા:$\frac{1 - (2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}) + (2 \cos^2 \frac{	heta}{2} - 1)}{1 - (2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}) - (1 - 2 \sin^2 \frac{	heta}{2})}$$= \frac{2 \cos^2 \frac{	heta}{2} - 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \sin^2 \frac{	heta}{2} - 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}$$= \frac{2 \cos \frac{	heta}{2} (\cos \frac{	heta}{2} - \sin \frac{	heta}{2})}{2 \sin \frac{	heta}{2} (\sin \frac{	heta}{2} - \cos \frac{	heta}{2})}$$= \frac{2 \cos \frac{	heta}{2} (\cos \frac{	heta}{2} - \sin \frac{	heta}{2})}{-2 \sin \frac{	heta}{2} (\cos \frac{	heta}{2} - \sin \frac{	heta}{2})}$$= -\frac{\cos \frac{	heta}{2}}{\sin \frac{	heta}{2}} = -\cot \frac{	heta}{2}$.

$\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta} = $

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