A) \[xyz+1=yz-zx\]
B) \[xyz-1=yz+zx\]
C) \[xyz-xy=yz-zx\]
D) \[xyz+1=yz+zx\]
Correct Answer: C
Solution :
[c]| Given, \[x=\sum\limits_{n=0}^{\infty }{{{\sin }^{2n}}\theta }\] \[=1+{{\sin }^{2}}\theta +{{\sin }^{4}}\theta +...\] |
| \[\Rightarrow \]\[x=\frac{1}{1-{{\sin }^{2}}\theta }=\frac{1}{{{\cos }^{2}}\theta }\] |
| \[\Rightarrow \]\[y=\sum\limits_{n=0}^{\infty }{{{\cos }^{2n}}\phi }\]\[=1+{{\cos }^{2}}\phi +{{\cos }^{4}}\phi +...\] |
| \[\Rightarrow y=\frac{1}{1-{{\cos }^{2}}\phi }=\frac{1}{{{\sin }^{2}}\phi }\] |
| \[z=\sum\limits_{n=0}^{\infty }{{{\cos }^{n}}\left( \theta +\phi \right)}{{\cos }^{n}}\left( \theta +\phi \right)\] |
| \[\Rightarrow \]\[z=1+\cos (\theta +\phi ){{\cos }^{2}}(\alpha -\phi )\]\[+{{\cos }^{2}}(\theta +\phi ){{\cos }^{2}}(\alpha -\phi )+...\] |
| \[\Rightarrow \]\[z=\frac{1}{1-\cos (\theta +\phi )\cos (\theta -\phi )}\] |
| \[\Rightarrow \]\[z=\frac{1}{1-\left( \frac{1}{x}-\frac{1}{y} \right)}=\frac{xy}{xy-(y-x)}\] |
| \[\Rightarrow \]\[xyz-(zy-zx)=xy\]\[\Rightarrow \]\[xyz-xy=yz-xz\] |
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