Please answer this question from Introduction to trigonometry.
if x = a tan θ and y = a sec θ, find the relation between x,

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1. Answer:

   [tex] {x}^{2} + {y}^{2} + {a}^{2} = 0[/tex]

   Step-by-step explanation:

   We have

   [tex]x = a \tan( \theta) \\ y = a \sec( \theta) [/tex]

   Squaring both equations,

   [tex] {x}^{2} = {a}^{2} \tan { \: }^{2} ( \theta) \\ {y}^{2} = {a}^{2} \sec { \: }^{2} ( \theta) [/tex]

   Adding both equations,

   [tex] {x}^{2} + {y}^{2} = {a}^{2} \tan {}^{2} ( \theta) + {a}^{2} \sec { }^{2} ( \theta) \\ {x}^{2} + {y}^{2} = {a}^{2} ( \tan {}^{2} ( \theta) + \sec {}^{2} ( \theta) )[/tex]

   Using

   [tex] \tan {}^{2} ( \theta) + \sec {}^{2} ( \theta) = – 1[/tex]

   [tex] {x}^{2} + {y}^{2} = {a}^{2} ( – 1) \\ {x}^{2} + {y}^{2} = – ( {a}^{2} ) \\ {x}^{2} + {y}^{2} + {a}^{2} = 0[/tex]

   Which is the required relation.

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