Question

if x=a sec y=b tan the the value of x2/a2 -y2/b2​

Answer 1

Step-by-step explanation:

imagine, x = 1 and y = 2

we know that, x = a and y = b

so,

x2/a2 = 12/12 = 1

-y2/b2 = -22/22 = -1

Answer:

x2/a2 = 1

-y2/b2 = -1

Answer 2

Answer:

1

Step-by-step explanation:

$x = a \sec(t) \\ y = b \tan(t)$

squaring these two equations and then putting in the formula we get

$\frac{x ^{2} }{ {a}^{2} } \: - \frac{y ^{2} }{b ^{2} }$

$\frac{ {a}^{2} \sec^{2} (t) }{a ^{2} } - \frac{ {b}^{2} \tan ^{2} (t) }{b ^{2} }$

${ \sec}^{2} t - {tan}^{2} t$

$= 1$

hence