STD 11 - 3. trignometrical ratios,functions and identities — JEE कक्षा 11 साइन्स — Question

$ = \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}}$