If x = a secθcos Φ, y= bsecθcos Φ, and z = ctanθ, show that x^2/a^2 + y^2/b^2 - z^2/c^2 = 1

Question

If x = a secθcos Φ, y= bsecθcos Φ, and z = ctanθ, show that \(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\) = 1

Answer 1

x = a secθcos Φ ⇒ x² = a² sec²θcos²Φ

∴ \(\frac{x^2}{a^2}\) = sec²θcos²Φ ......(1)

y= bsecθcos Φ ⇒ y²= b²sec²θsin²Φ

∴ \(\frac{y^2}{a^2}\) sec²θsin²Φ ...(2)

z = ctanθ ⇒ z² = c²tan²θ 

∴ \(\frac{z^2}{c^2}\) = tan²θ   ......(3)

Now, \(\frac{x^2}{a^2}+\frac{y^2}{a^2}-\frac{z^2}{c^2}\) = sec²θcos²Φ + sec²θsin²Φ - tan²θ

 = sec²θ(cos²Φ + sin²Φ) - tan²θ

 = sec²θ x 1 - tan²​​​​​​​θ

=  sec²θ - tan²​​​​​​​θ = 1

Hence , \(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\) = 1

Hence Proved.