| If\[{{\cos }^{4}}x+{{\cos }^{2}}x=1,\] then the value of \[{{\tan }^{4}}x+{{\tan }^{2}}x\]is [SSC (CPO) 2013] |
A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[1\]
D) 2
Correct Answer: C
Solution :
| Given, \[{{\cos }^{4}}x+{{\cos }^{2}}x=1\] |
| \[\Rightarrow \]\[{{\cos }^{4}}x=1-{{\cos }^{2}}x\]\[[\because {{\sin }^{2}}x+{{\cos }^{2}}x=1]\] |
| \[\Rightarrow \]\[{{\cos }^{4}}x={{\sin }^{2}}x\] (i) |
| Now, \[{{\tan }^{4}}x+{{\tan }^{2}}x=?\] |
| [To find the value of \[({{\tan }^{4}}x+{{\tan }^{2}}x),\]we have to convert it into\[\sin \theta \]and \[\cos \theta \]form] |
| Hence, \[\frac{{{\sin }^{4}}x}{{{\cos }^{4}}x}+\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\] \[\left[ \because \tan \theta =\frac{\sin \theta }{\cos \theta } \right]\] |
| \[=\frac{{{\sin }^{4}}x+{{\sin }^{2}}x{{\cos }^{2}}x}{{{\cos }^{4}}x}\] |
| \[=\frac{{{\sin }^{2}}x({{\sin }^{2}}x+{{\cos }^{2}}x)}{{{\cos }^{4}}x}\] |
| \[=\frac{{{\sin }^{2}}x\times 1}{{{\sin }^{2}}x}=1\] |
| [from Eq. (i),\[{{\cos }^{4}}x={{\sin }^{2}}x\]] |
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