If x = sec, phi - tan, phi & y = cosec, p | vedclass

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Question

$x = \frac{{1 - \sin \phi }}{{\cos \phi }}\,\, = \,\,\frac{{1 - \cos (\pi /2 - \phi )}}{{\sin (\pi /2 - \phi )}}$ $= tan(\pi /4-\phi /2)$

$y =

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$x = \frac{{1 - \sin \phi }}{{\cos \phi }}\,\, = \,\,\frac{{1 - \cos (\pi /2 - \phi )}}{{\sin (\pi /2 - \phi )}}$ $= tan(\pi /4-\phi /2)$

$y = \frac{{1 + \cos \phi }}{{\sin \phi }}\,\, = \,\,\frac{{2\,{{\cos }^2}\phi /2}}{{2\sin (\phi /2)\,\cos (\phi /2)}}$ $= cot\, \phi /2$

$x = \frac{{1 - 	an \phi /2}}{{1 + 	an \phi /2}}$ $=$$\frac{{\cot \phi /2\, - \,1}}{{\cot \phi /2\, + 1}}\,\, = \,\,\frac{{y - 1}}{{y + 1}}$

Applying $C/D$ $\frac{{x + 1}}{{x - 1}}\,\, = \,\,\frac{{y - 1 + y + 1}}{{y - 1 - y - 1}}\,\, = \,\,y$

Also ,$y =$$\frac{{x + 1}}{{1 - x}}\,\, \Rightarrow $ $y - xy - x-1 = 0$

If $x = sec\, \phi - tan\, \phi$ & $y = cosec\, \phi + cot\, \phi$ then :

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