If x= a secθ cos Φ , y = b secθ sin Φ z = c tan θ, then, the value of $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}$$ is :

$$x= a secθ cos Φ$$
$$x^2 = a^2 sec^2θ cos^2Φ $$
$$ \frac{x^2}{a^2} = sec^2θ cos^2Φ $$

$$y= b secθ sin Φ$$
$$y^2 = b^2 sec^2θ sin^2Φ $$
$$ \frac{y^2}{b^2} = sec^2θ sin^2Φ $$

$$z= c tanθ $$
$$z^2 = c^2 tan^2θ $$
$$ \frac{z^2}{c^2} = tan^2θ $$

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}} = sec^2θ cos^2Φ + sec^2θ sin^2Φ - tan^2θ$$
$$sec^2θ cos^2Φ + sec^2θ sin^2Φ - tan^2θ = sec^2θ (cos^2Φ + sin^2Φ)- tan^2θ$$
=$$sec^2θ - tan^2θ = 1$$
Option A is the correct answer.