A) \[\left( 1+\frac{{{z}^{2}}}{{{c}^{2}}} \right)\]
B) \[\left( 1-\frac{{{z}^{2}}}{{{c}^{2}}} \right)\]
C) \[\left( \frac{{{z}^{2}}}{{{c}^{2}}}-1 \right)\]
D) \[\frac{{{z}^{2}}}{{{c}^{2}}}\]
Correct Answer: A
Solution :
| [a] We have, \[\sec \theta \cos \phi =\frac{x}{a}\]...(1) |
| \[\sec \theta \sin \phi =\frac{y}{b}\] (2) |
| and \[\tan \theta =\frac{z}{c}\] ...(3) |
| Squaring and adding eqs. (1) and (2), we get |
| \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}={{\sec }^{2}}\theta {{\cos }^{2}}\phi +{{\sec }^{2}}\theta {{\sin }^{2}}\phi \] |
| \[={{\sec }^{2}}\theta ({{\cos }^{2}}\phi +{{\sin }^{2}}\phi )={{\sec }^{2}}\theta \] \[[{{\cos }^{2}}A+{{\sin }^{2}}A=1]\] |
| \[=(1+{{\tan }^{2}}\theta )=\left( 1+\frac{{{z}^{2}}}{{{c}^{2}}} \right)\] [From eq.(3)] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\left( \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}} \right)=\left( 1+\frac{{{z}^{2}}}{{{c}^{2}}} \right)\] |
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