Question: If $ {{\cos }^{4}}x+{{\cos }^{2}}x=1, $ then the value of $ {{\tan }^{4}}x+{{\tan }^{2}}x $ is
Options:
A) $ \frac{1}{4} $
B) $ \frac{1}{2} $
C) $ 1 $
D) 2
Show Answer
Answer:
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} $ $ [ \because \tan \theta =\frac{\sin \theta }{\cos \theta } ] $
$ =\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 $ ]
BETA
