If x = sec ,phi - tan phi ,y = {rm{cosec | vedclass

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Question

(b) We have $xy = (\sec \phi- 	an \phi)\,\,{m{(cosec}}\,\,\phi+ \cot \,\,\phi)$ 

$ = \frac{{1 - \sin \,\phi}}{{\cos \,\phi}}\,.\,\frac{{1 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) We have $xy = (\sec \phi- 	an \phi)\,\,{m{(cosec}}\,\,\phi+ \cot \,\,\phi)$ 

$ = \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}}$

$= \frac{{1 - \sin \,\phi+ \cos \,\phi}}{{\cos \,\phi\sin \,\phi}}$&hellip;..$(i)$ 

$x - y = (\sec \,\phi- 	an \,\phi) - (\cos ec\,\phi+ \cot \,\phi)$ $ = \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}}$&hellip;..$(ii)$

Adding $(i)$ and $(ii)$ we get, $xy + 1 + (x - y) = 0$ 

$ \Rightarrow x = \frac{{y - 1}}{{y + 1}}$.

If $x = \sec \,\phi - \tan \phi ,y = {\rm{cosec}}\phi + \cot \phi ,$ then

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