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

$\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}=$
  1. $\cot \frac{\theta}{2}$
  2. $-\cot \frac{\theta}{2}$
  3. $\tan \frac{\theta}{2}$
  4. $-\tan \frac{\theta}{2}$

Solution

$\begin{array}{l} \text { We know } \sin \theta=2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2} \text { and } \cos \theta=2 \cos ^{2} \frac{\theta}{2}-1=1-2 \sin ^{2} \frac{\theta}{2} \\ \frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}=\frac{1-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}+\left(2 \cos ^{2} \frac{\theta}{2}-1\right)}{1-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}-\left(1-2 \sin ^{2} \frac{\theta}{2}\right)} \end{array}$ $=\frac{-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}+2 \cos ^{2} \frac{\theta}{2}}{-2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}+2 \sin ^{2} \frac{\theta}{2}}$ $=\frac{-2 \cos \frac{\theta}{2}\left(\sin \frac{\theta}{2}-\cos \frac{\theta}{2}\right)}{-2 \sin \frac{\theta}{2}\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)}=-\cot \frac{\theta}{2}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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