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यदि $x = \sec \,\phi - \tan \phi ,y = {\rm{cosec}}\phi + \cot \phi ,$ तो
$x = \frac{{y + 1}}{{y - 1}}$
$x = \frac{{y - 1}}{{y + 1}}$
$y = \frac{{1 - x}}{{1 + x}}$
इनमें से कोई नहीं
Solution
दिया है $xy = (\sec \phi – \tan \phi )\,\,{\rm{(cosec}}\,\,\phi + \cot \,\,\phi )$
$ = \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)$
$(i)$ व $(ii)$ को जोड़ने पर, $xy + 1 + (x – y) = 0$
$ \Rightarrow x = \frac{{y – 1}}{{y + 1}}$
