If a<sup>2</sup> sec<sup>2</sup>x - b<sup>2</sup> tan<sup>2</sup>x = c<sup>2</sup>, then the value of sec<sup>2</sup>x + tan<sup>2</sup>x is equal to (assume b<sup>2</sup> ≠ a<sup>2</sup>)
Correct Answer: $$\frac{{{b^2} + {a^2} - 2{c^2}}}{{{b^2} - {a^2}}}$$
$$\eqalign{ & {a^2}{\sec ^2}x - {b^2}{\tan ^2}x = {c^2} \cr & \Rightarrow {a^2}\left( {1 + ta{n^2}x} \right) - {b^2}{\tan ^2}x = {c^2} \cr & \Rightarrow {a^2} + {a^2}{\tan ^2}x - {b^2}{\tan ^2}x = {c^2} \cr & \Rightarrow {a^2} + ta{n^2}x\left( {{a^2} - {b^2}} \right) = {c^2} \cr & \Rightarrow {a^2} - {c^2} = {\tan ^2}x\left( {{b^2} - {a^2}} \right) \cr & \Rightarrow \frac{{{a^2} - {c^2}}}{{{b^2} + {a^2}}} = {\text{ta}}{{\text{n}}^2}x \cr & \Rightarrow {\sec ^2}x - {\tan ^2}x = 1 \cr & \Rightarrow {\sec ^2}x = {\tan ^2}x + 1 \cr & \Rightarrow 1 + \frac{{{a^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & \Rightarrow \frac{{{b^2} - {a^2} + {a^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & \Rightarrow \frac{{{b^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & {\sec ^2}x + {\tan ^2}x \cr & = \frac{{{b^2} - {c^2}}}{{{b^2} - {a^2}}} + \frac{{{a^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & = \frac{{{b^2} + {a^2} - 2{c^2}}}{{{b^2} - {a^2}}} \cr} $$