# parametric equation in trigonometric

• Apr 11th 2007, 03:46 AM
T_o_M
parametric equation in trigonometric
if x = a sin c and y = b tan c show that (a^2)/(x^2) - (b^2)/(y^2) = 1

where x, y, a, b and c are just pronumerals...

• Apr 11th 2007, 03:55 AM
DivideBy0
Let's write this out fully:

a^2 / (a*sin(c))^2 - b^2 / (b*sin(c))^2 = 1
a^2 / (a^2*sin^2(c)) - b^2 / (b^2*sin^2(c)) = 1

The a^2 and b^2's cancel out

1/sin^2(c) - 1/tan^2(c) = 1
1/sin^2(c) - cos^2(c)/sin^2(c) = 1

Under the same denominator...

(1 - cos^2(c)) / sin^2(c) = 1

Since we know sin^2(c) + cos^2(c) = 1 (Pythagorean identity),
1 - cos^2(c) = sin^2(c)

Thus, we have :

sin^2(c) / sin^2(c) = 1
1=1

:)
• Apr 11th 2007, 07:33 AM
Soroban
Hello, T_o_M!

Quote:

If x = a·sinθ and y = b·tanθ, show that: .a²/x² - b²/y² .= .1

x = a·sinθ . . cscθ = a/x .[1]

y = b·tanθ . . cotθ = b/y .[2]

Square [1]: .a²/x² .= .csc²θ
Square [2]: .b²/y² .= .cot²θ

Subtract: .a²/x² - b²/y² .= .csc²θ - cot²θ

Since csc²θ - cot²θ = 1, we have: .a²/x² - b²/y² .= .1