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Prove that:$ \frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1} =\operatorname{cosec} \theta+\cot \theta $
To do:
We have to prove that \( \frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1} =\operatorname{cosec} \theta+\cot \theta \).
Solution:
We know that,
$\sin^2 A+\cos^2 A=1$
$\operatorname{cosec}^2 A-\cot^2 A=1$
$\sec^2 A-\tan^2 A=1$
$\cot A=\frac{\cos A}{\sin A}$
$\tan A=\frac{\sin A}{\cos A}$
$\operatorname{cosec} A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Therefore,
$\frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1}=\frac{\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\sin \theta}+\frac{1}{\sin \theta}}{\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\sin \theta}-\frac{1}{\sin \theta}}$ (Divide each term by $\sin \theta$)
$=\frac{\cot \theta-1+\operatorname{cosec} \theta}{\cot \theta+1-\operatorname{cosec} \theta}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta-1)}{(\cot \theta-\operatorname{cosec} \theta+1)}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta-1)(\cot \theta+\operatorname{cosec} \theta)}{[(\cot \theta-\operatorname{cosec} \theta)+1](\cot \theta+\operatorname{cosec} \theta)}$ (Multiplying and dividing by $\cot \theta+\operatorname{cosec} \theta$)
$=\frac{(\cot \theta+\operatorname{cosec} \theta-1)(\cot \theta+\operatorname{cosec} \theta)}{(\cot \theta-\operatorname{cosec} \theta)(\cot \theta+\operatorname{cosec} \theta)+\cot \theta+\operatorname{cosec} \theta}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta-1)(\cot \theta+\operatorname{cosec} \theta)}{\left(\cot ^{2} \theta-\operatorname{cosec}^{2} \theta\right)+\cot \theta+\operatorname{cosec} \theta}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta-1)(\cot \theta+\operatorname{cosec} \theta)}{-1+\cot \theta+\operatorname{cosec} \theta}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta-1)(\cot \theta+\operatorname{cosec} \theta)}{(\cot \theta+\operatorname{cosec} \theta-1)}$
$=\cot \theta+\operatorname{cosec} \theta$
$=\operatorname{cosec} \theta+\cot \theta$
Hence proved.
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