Question

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

Answer 1

Answer:

${x}^{2} + {y}^{2} + {a}^{2} = 0$

Step-by-step explanation:

We have

$x = a \tan( \theta) \\ y = a \sec( \theta)$

Squaring both equations,

${x}^{2} = {a}^{2} \tan { \: }^{2} ( \theta) \\ {y}^{2} = {a}^{2} \sec { \: }^{2} ( \theta)$

Adding both equations,

${x}^{2} + {y}^{2} = {a}^{2} \tan {}^{2} ( \theta) + {a}^{2} \sec { }^{2} ( \theta) \\ {x}^{2} + {y}^{2} = {a}^{2} ( \tan {}^{2} ( \theta) + \sec {}^{2} ( \theta) )$

Using

$\tan {}^{2} ( \theta) + \sec {}^{2} ( \theta) = - 1$

${x}^{2} + {y}^{2} = {a}^{2} ( - 1) \\ {x}^{2} + {y}^{2} = - ( {a}^{2} ) \\ {x}^{2} + {y}^{2} + {a}^{2} = 0$

Which is the required relation.