If x = a sec $\theta$ cos ϕ, y = b sec $\theta$ sin ϕ and z = c tan $\theta$, thenx^2/a^2 + y^2/b^2

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Given: x= a sec$\theta$cosΦ $\Rightarrow$ x/a=sec$\theta$ cosΦ y=b sec $\theta$ sinΦ $\Rightarrow$ y/b=sec$\theta$ sinΦ z=c tan $\theta$ z/c= tan $\theta$ Now, `(x/a)^2+(y/b)^2-(z/c)^2=(sec$\theta$ cosΦ)^2+(sec$\theta$ sin Φ)^2-(tan$\theta$ )^2 $\Rightarrow$ x^2/a^2+y^2/b^2-z^2/c^2= sec^2$\theta$cos^2 Φ+sec^2$\theta$sin^2Φ-tan^2$\theta$ $\Rightarrow$ x^2/a^2+y^2/b^2-z^2/c^2=(sec^2$\theta$ cos^2Φ+sec^2$\theta$ sin^2 sin^2Φ)-tan^2Φ $\Rightarrow$ x^2/a^2+y^2/b^2-z^2/c^2=sec^2$\theta$(cos^2Φ+sin^2Φ)-tan^2$\theta$ $\Rightarrow$ x^2/a^2+y^2/b^2-z^2/c^2= sec^2$\theta$(1)-tan^2$\theta$ $\Rightarrow$ x^2/a^2+y^2/b^2-z^2/c^2=sec^2$\theta$-tan^2$\theta$ $\Rightarrow$ x^2/a^2+y^2/b^2-z^2/c^2=1 $\Rightarrow$x^2/a^2+y^2/b^2=1+z^2/c^2`

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