If \[ax+b(sec(ta{{n}^{-1}}x))=c\] and \[ay+b\]\[(sec\,.\,(ta{{n}^{-1}}y))=c,\] then \[\frac{x+y}{1-xy}=\]

\[\frac{ac}{{{a}^{2}}+{{c}^{2}}}\]

\[\frac{2ac}{{{a}^{2}}-{{c}^{2}}}\]

If \[ax+b(sec(ta{{n}^{-1}}x))=c\] and \[ay+b\]\[(sec\,.\,(ta{{n}^{-1}}

1.	If \[ax+b(sec(ta{{n}^{-1}}x))=c\] and \[ay+b\]\[(sec\,.\,(ta{{n}^{-1}}y))=c,\] then \[\frac{x+y}{1-xy}=\]
A.	\[\frac{ac}{{{a}^{2}}+{{c}^{2}}}\]
B.	\[\frac{2ac}{a-c}\]
C.	\[\frac{2ac}{{{a}^{2}}-{{c}^{2}}}\]
D.	\[\frac{a+c}{1-ac}\]
Answer» D. \[\frac{a+c}{1-ac}\]

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If \[ax+b(sec(ta{{n}^{-1}}x))=c\] and \[ay+b\]\[(sec\,.\,(ta{{n}^{-1}}y))=c,\] then \[\frac{x+y}{1-xy}=\]

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