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If $x = \sec \,\phi - \tan \phi ,y = {\rm{cosec}}\phi + \cot \phi ,$ then
$x = \frac{{y + 1}}{{y - 1}}$
$x = \frac{{y - 1}}{{y + 1}}$
$y = \frac{{1 - x}}{{1 + x}}$
None of these
Solution
(b) We have $xy = (\sec \phi- \tan \phi)\,\,{\rm{(cosec}}\,\,\phi+ \cot \,\,\phi)$
$ = \frac{{1 – \sin \,\phi}}{{\cos \,\phi}}\,.\,\frac{{1 + \cos \,\phi}}{{\sin \,\phi}}$
$ \Rightarrow \,xy + 1 = \frac{{1 – \sin \,\phi+ \cos \,\phi- \sin \,\phi\,\cos \,\phi+ \sin \phi \cos \phi}}{{\cos \phi\sin \phi}}$
$= \frac{{1 – \sin \,\phi+ \cos \,\phi}}{{\cos \,\phi\sin \,\phi}}$…..$(i)$
$x – y = (\sec \,\phi- \tan \,\phi) – (\cos ec\,\phi+ \cot \,\phi)$ $ = \frac{{1 – \sin \,\phi}}{{\cos \,\phi}} – \frac{{1 + \cos \,\phi}}{{\sin \,\phi}}$
$= \frac{{\sin \,\phi- {{\sin }^2}\phi- \cos \,\phi- {{\cos }^2}\phi}}{{\cos \,\phi\,\sin \,\phi}}$
$ = \frac{{\sin \,\phi – \cos \,\phi- 1}}{{\cos \,\phi \,\sin \,\phi}}$…..$(ii)$
Adding $(i)$ and $(ii)$ we get, $xy + 1 + (x – y) = 0$
$ \Rightarrow x = \frac{{y – 1}}{{y + 1}}$.