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 $x

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $x = \frac{1 - \sin \phi}{\cos \phi}$. Then $\frac{1}{x} = \frac{\cos \phi}{1 - \sin \phi} = \frac{\cos \phi(1 + \sin \phi)}{1 - \sin^2 \phi} = \frac{\cos \phi(1 + \sin \phi)}{\cos^2 \phi} = \frac{1 + \sin \phi}{\cos \phi}$.Similarly,$y = \frac{1 + \cos \phi}{\sin \phi}$. Then $\frac{1}{y} = \frac{\sin \phi}{1 + \cos \phi} = \frac{\sin \phi(1 - \cos \phi)}{1 - \cos^2 \phi} = \frac{\sin \phi(1 - \cos \phi)}{\sin^2 \phi} = \frac{1 - \cos \phi}{\sin \phi}$.Alternatively,using the identity $\sec \phi - 	an \phi = 	an(\frac{\pi}{4} - \frac{\phi}{2})$ and $\csc \phi + \cot \phi = \cot(\frac{\phi}{2})$,we can relate $x$ and $y$. Let's test the relation $x = \frac{y - 1}{y + 1}$.Substituting $y = \frac{1 + \cos \phi}{\sin \phi}$ into $\frac{y - 1}{y + 1}$:$\frac{\frac{1 + \cos \phi}{\sin \phi} - 1}{\frac{1 + \cos \phi}{\sin \phi} + 1} = \frac{1 + \cos \phi - \sin \phi}{1 + \cos \phi + \sin \phi} = \frac{2\cos^2(\phi/2) - 2\sin(\phi/2)\cos(\phi/2)}{2\cos^2(\phi/2) + 2\sin(\phi/2)\cos(\phi/2)} = \frac{\cos(\phi/2) - \sin(\phi/2)}{\cos(\phi/2) + \sin(\phi/2)} = 	an(\frac{\pi}{4} - \frac{\phi}{2}) = x$.Thus,$x = \frac{y - 1}{y + 1}$ is correct.

If $x = \sec \phi - \tan \phi$ and $y = \csc \phi + \cot \phi$,then which of the following is true?

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