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$\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}=$
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Verified Answer
The correct answer is:
$-\cot \frac{\theta}{2}$
(D)
$\begin{array}{l}
\text { We know } \sin \theta=2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2} \text { and } \cos \theta=2 \cos ^{2} \frac{\theta}{2}-1=1-2 \sin ^{2} \frac{\theta}{2} \\
\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}=\frac{1-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}+\left(2 \cos ^{2} \frac{\theta}{2}-1\right)}{1-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}-\left(1-2 \sin ^{2} \frac{\theta}{2}\right)}
\end{array}$
$=\frac{-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}+2 \cos ^{2} \frac{\theta}{2}}{-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}+2 \sin ^{2} \frac{\theta}{2}}$
$=\frac{-2 \cos \frac{\theta}{2}\left(\sin \frac{\theta}{2}-\cos \frac{\theta}{2}\right)}{-2 \sin \frac{\theta}{2}\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)}=-\cot \frac{\theta}{2}$
$\begin{array}{l}
\text { We know } \sin \theta=2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2} \text { and } \cos \theta=2 \cos ^{2} \frac{\theta}{2}-1=1-2 \sin ^{2} \frac{\theta}{2} \\
\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}=\frac{1-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}+\left(2 \cos ^{2} \frac{\theta}{2}-1\right)}{1-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}-\left(1-2 \sin ^{2} \frac{\theta}{2}\right)}
\end{array}$
$=\frac{-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}+2 \cos ^{2} \frac{\theta}{2}}{-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}+2 \sin ^{2} \frac{\theta}{2}}$
$=\frac{-2 \cos \frac{\theta}{2}\left(\sin \frac{\theta}{2}-\cos \frac{\theta}{2}\right)}{-2 \sin \frac{\theta}{2}\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)}=-\cot \frac{\theta}{2}$
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