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

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(B) Given $x = \sec \phi - 	an \phi = \frac{1 - \sin \phi}{\cos \phi}$ and $y = \csc \phi + \cot \phi = \frac{1 + \cos \phi}{\sin \phi}$.Consider the

Vedclass · Accepted Answer

$x = \frac{y - 1}{y + 1}$

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(B) Given $x = \sec \phi - 	an \phi = \frac{1 - \sin \phi}{\cos \phi}$ and $y = \csc \phi + \cot \phi = \frac{1 + \cos \phi}{\sin \phi}$.Consider the expression $\frac{y - 1}{y + 1}$:$y - 1 = \frac{1 + \cos \phi}{\sin \phi} - 1 = \frac{1 + \cos \phi - \sin \phi}{\sin \phi}$$y + 1 = \frac{1 + \cos \phi}{\sin \phi} + 1 = \frac{1 + \cos \phi + \sin \phi}{\sin \phi}$Therefore,$\frac{y - 1}{y + 1} = \frac{1 + \cos \phi - \sin \phi}{1 + \cos \phi + \sin \phi}$.Multiply the numerator and denominator by $(1 + \cos \phi - \sin \phi)$:$= \frac{(1 + \cos \phi - \sin \phi)^2}{(1 + \cos \phi)^2 - \sin^2 \phi} = \frac{1 + \cos^2 \phi + \sin^2 \phi + 2\cos \phi - 2\sin \phi - 2\sin \phi \cos \phi}{1 + \cos^2 \phi + 2\cos \phi - \sin^2 \phi}$$= \frac{2 + 2\cos \phi - 2\sin \phi - 2\sin \phi \cos \phi}{2\cos^2 \phi + 2\cos \phi} = \frac{2(1 + \cos \phi)(1 - \sin \phi)}{2\cos \phi(1 + \cos \phi)}$$= \frac{1 - \sin \phi}{\cos \phi} = x$.Thus,$x = \frac{y - 1}{y + 1}$.

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

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