If x = a sec A cos B, y = b sec A sin B and z = c tan A, prove that $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1$

⇒ x = a sec A cos B

sec A cos B = $\dfrac{x}{a}$….(1)

⇒ y = b sec A sin B

sec A sin B = $\dfrac{y}{b}$….(2)

Square and add equations (1) and (2) :

$\Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \sec^2 A \cos^2 B + \sec^2 A \sin^2 B \\[1em] \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \sec^2 A (\cos^2 B + \sin^2 B) \\[1em] \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \sec^2 A …(3)$

Now,

⇒ z = c tan A

tan A = $\dfrac{z}{c}$

tan² A = $\dfrac{z^2}{c^2}$…(4)

Subtract (4) from (3):

$\Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = \sec^2 A - \tan^2 A \\[1em] \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1.$

Hence, proved that $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1$.