A) \[\frac{1}{4}\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[0\]
D) \[1\]
Correct Answer: D
Solution :
\[{{\cos }^{4}}x+{{\cos }^{2}}x=1\] \[{{\cos }^{4}}x=1-{{\cos }^{2}}x-{{\cos }^{4}}x={{\sin }^{2}}x\] Divide both side by \[{{\cos }^{2}}x\] \[{{\cos }^{2}}x=\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}={{\cos }^{2}}x={{\tan }^{2}}x\] Now, \[{{\tan }^{4}}x+{{\tan }^{2}}x={{({{\tan }^{2}}x)}^{2}}+{{\tan }^{2}}x\] \[=\,\,\,\,{{\cos }^{4}}x+{{\cos }^{2}}x=\underline{\mathbf{1}}\]
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