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...

please answer

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- April 11th 2007, 02:46 AMT_o_Mparametric 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...

please answer - April 11th 2007, 02:55 AMDivideBy0
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

:) - April 11th 2007, 06:33 AMSoroban
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