Quantitative Aptitude Ques 2428
Question: The value of $ (2{{\cos }^{2}}\theta -1)( \frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta } ) $
Options:
A) 4
B) 1
C) 3
D) 2
Show Answer
Answer:
Correct Answer: C
Solution:
- $ (2{{\cos }^{2}}\theta -1)( \frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta } ) $ $ =(2{{\cos }^{2}}\theta -1)( \frac{{{(1+\tan \theta )}^{2}}+{{(1-tan\theta )}^{2}}}{1-{{\tan }^{2}}\theta } ) $
$ =(2{{\cos }^{2}}\theta -1)[ \frac{ \begin{aligned} & (1^{2}+{{\tan }^{2}}\theta +2\tan \theta ) \\ & +,(1^{2}+{{\tan }^{2}}\theta -2\tan \theta \\ \end{aligned}}{1-\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }} ] $
$ =(2{{\cos }^{2}}\theta -1)[ \frac{ \begin{aligned} & (1+{{\tan }^{2}}\theta +2\tan \theta ) \\ & +,(1+{{\tan }^{2}}\theta -2\tan \theta \\ \end{aligned}}{\frac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }} ] $
$ =(2{{\cos }^{2}}\theta -1)[ \frac{2,(1+{{\tan }^{2}}\theta ).{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta } ] $
$ =(2{{\cos }^{2}}\theta -1),\frac{2{{\sec }^{2}}\theta .{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta ,-(1-{{\cos }^{2}}\theta )} $
$ [\because 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta ] $
$ =(2{{\cos }^{2}}\theta -1)\frac{2{{\sec }^{2}}\theta \frac{1}{{{\sec }^{2}}\theta }}{2{{\cos }^{2}}\theta -1}=2 $
BETA
