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If $\mathrm{x}=\mathrm{a} \sec \theta \cos \phi, \mathrm{y}=\mathrm{bsec} \theta \sin \phi, \mathrm{z}=\operatorname{ctan} \theta$, then what
is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}$ equal to?
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is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}$ equal to?
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1906 Upvotes
Verified Answer
The correct answer is:
1
As given:
$\mathrm{x}=\mathrm{a} \sec \theta \cos \phi, \mathrm{y}=\mathrm{b} \sec \theta \sin \phi, \mathrm{z}=\mathrm{c} \tan \theta$
So, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=\frac{a^{2} \sec ^{2} \theta \cos ^{2} \phi}{a^{2}}$
$+\frac{b^{2} \sec ^{2} \theta \sin ^{2} \phi}{b^{2}}-\frac{c^{2} \tan ^{2} \theta}{c^{2}}$
$=\sec ^{2} \theta\left(\cos ^{2} \phi+\sin ^{2} \phi\right)-\tan ^{2} \theta=\sec ^{2} \theta-\tan ^{2} \theta=1$
$\mathrm{x}=\mathrm{a} \sec \theta \cos \phi, \mathrm{y}=\mathrm{b} \sec \theta \sin \phi, \mathrm{z}=\mathrm{c} \tan \theta$
So, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=\frac{a^{2} \sec ^{2} \theta \cos ^{2} \phi}{a^{2}}$
$+\frac{b^{2} \sec ^{2} \theta \sin ^{2} \phi}{b^{2}}-\frac{c^{2} \tan ^{2} \theta}{c^{2}}$
$=\sec ^{2} \theta\left(\cos ^{2} \phi+\sin ^{2} \phi\right)-\tan ^{2} \theta=\sec ^{2} \theta-\tan ^{2} \theta=1$
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