If x = sec phi - tan phi and y = csc phi + co | vedclass

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(D) Given $x = \sec \phi - 	an \phi = \frac{1 - \sin \phi}{\cos \phi} = \frac{1 - \cos(\frac{\pi}{2} - \phi)}{\sin(\frac{\pi}{2} - \phi)} = 	an(\fra

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All of the above

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(D) Given $x = \sec \phi - 	an \phi = \frac{1 - \sin \phi}{\cos \phi} = \frac{1 - \cos(\frac{\pi}{2} - \phi)}{\sin(\frac{\pi}{2} - \phi)} = 	an(\frac{\pi}{4} - \frac{\phi}{2})$.Given $y = \csc \phi + \cot \phi = \frac{1 + \cos \phi}{\sin \phi} = \frac{2 \cos^2(\frac{\phi}{2})}{2 \sin(\frac{\phi}{2}) \cos(\frac{\phi}{2})} = \cot(\frac{\phi}{2})$.Now,$x = 	an(\frac{\pi}{4} - \frac{\phi}{2}) = \frac{	an(\frac{\pi}{4}) - 	an(\frac{\phi}{2})}{1 + 	an(\frac{\pi}{4}) 	an(\frac{\phi}{2})} = \frac{1 - 	an(\frac{\phi}{2})}{1 + 	an(\frac{\phi}{2})}$.Since $y = \cot(\frac{\phi}{2}) = \frac{1}{	an(\frac{\phi}{2})}$,we have $	an(\frac{\phi}{2}) = \frac{1}{y}$.Substituting this into $x$,we get $x = \frac{1 - \frac{1}{y}}{1 + \frac{1}{y}} = \frac{y - 1}{y + 1}$.This confirms option $C$ is correct.From $x = \frac{y - 1}{y + 1}$,we get $x(y + 1) = y - 1$ $\Rightarrow xy + x = y - 1$ $\Rightarrow xy + x - y + 1 = 0$. This confirms option $A$ is correct.Also,$y(1 - x) = 1 + x \Rightarrow y = \frac{1 + x}{1 - x}$. This confirms option $B$ is correct.Therefore,all options are correct.

If $x = \sec \phi - \tan \phi$ and $y = \csc \phi + \cot \phi$,then:

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