If frac{2 sin alpha}{1+cos alpha+sin alpha}=x | vedclass

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(B) Given $\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=x$. Multiply the numerator and denominator by $(1-(\cos \alpha+\sin \alpha))$: $\frac{2 \si

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$-x$

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(B) Given $\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=x$. Multiply the numerator and denominator by $(1-(\cos \alpha+\sin \alpha))$: $\frac{2 \sin \alpha(1-\cos \alpha-\sin \alpha)}{(1+\cos \alpha+\sin \alpha)(1-(\cos \alpha+\sin \alpha))}=x$ $\Rightarrow \frac{2 \sin \alpha(1-\cos \alpha-\sin \alpha)}{1^2-(\cos \alpha+\sin \alpha)^2}=x$ $\Rightarrow \frac{2 \sin \alpha(1-\cos \alpha-\sin \alpha)}{1-(\cos^2 \alpha+\sin^2 \alpha+2 \cos \alpha \sin \alpha)}=x$ Using $\sin^2 \alpha+\cos^2 \alpha=1$: $\Rightarrow \frac{2 \sin \alpha(1-\cos \alpha-\sin \alpha)}{1-(1+2 \cos \alpha \sin \alpha)}=x$ $\Rightarrow \frac{2 \sin \alpha(1-\cos \alpha-\sin \alpha)}{-2 \cos \alpha \sin \alpha}=x$ $\Rightarrow \frac{1-\cos \alpha-\sin \alpha}{-\cos \alpha}=x$ $\Rightarrow \frac{1-\cos \alpha-\sin \alpha}{\cos \alpha}=-x$

If $\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=x$,then $\frac{1-\cos \alpha-\sin \alpha}{\cos \alpha}=$

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