| If \[x=a\,\,(\sin \theta +\cos \theta ),\]\[y=b\,\,(\sin \theta -\cos \theta ),\]then the value of \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}\]is [SSC (CGL) Mains 2014] |
A) \[0\]
B) \[1\]
C) \[2\]
D) \[-\,\,2\]
Correct Answer: C
Solution :
| Given, \[x=a\,\,(\sin \theta +cos\theta )\] |
| On squaring both sides, we get |
| \[{{x}^{2}}={{a}^{2}}\,\,({{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta )\] |
| \[={{a}^{2}}(1+2\sin \theta \cos \theta )\] |
| Similarly, \[y=b\,\,(\sin \theta -\cos \theta )\] |
| \[\Rightarrow \]\[{{y}^{2}}={{b}^{2}}\,\,({{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta )\] |
| \[={{b}^{2}}\,\,(1-2\sin \theta \cos \theta )\] |
| Now, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=\frac{{{a}^{2}}\,\,(1+2\sin \theta \cos \theta )}{{{a}^{2}}}\] |
| \[+\frac{{{b}^{2}}(1-2\sin \theta \cos \theta )}{{{b}^{2}}}\] |
| \[=1+2\sin \theta \cos \theta +1-2\sin \theta \cos \theta \] |
| \[=2\] |
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