JEE Main Solved Paper 2017 Question 21
Question: $ 5({{\tan }^{2}}x-{{\cos }^{2}}x)=2\cos 2x+9, $ then the value of $ \cos 4x $ is :- [JEE Main Solved Paper-2017]
Options:
A) $ -\frac{7}{9} $
B) $ -\frac{3}{5} $
C) $ \frac{1}{3} $
D) $ \frac{2}{9} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ 5[ \frac{1-t}{t}-t ]=2(2t-1)+9 $ $ {Let,{{\cos }^{2}}x=t} $
$ \Rightarrow $ $ 5(1-t-t^{2})=t(4t+7) $
$ \Rightarrow $ $ 9t^{2}+12t-5=0 $
$ \Rightarrow $ $ 9t^{2}+15t-3t-5=0 $
$ \Rightarrow $ $ (3t-1)(3t+5)=0 $
$ \Rightarrow $ $ t=\frac{1}{3}as,t\ne -\frac{5}{3}. $ $ \cos 2x=2( \frac{1}{3} )-1=-\frac{1}{3} $ $ \cos 4x=2{{( -\frac{1}{3} )}^{2}}-1=-\frac{7}{9} $
