3. trignometrical ratios,functions and identities — ગણિત ધોરણ 11 સાયન્સ — 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)$

Adding $(i)$ and $(ii)$ we get,

$xy + 1 + (x - y) = 0$

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