Inverse Trigonometric Functions Question 167
Question: $ \tan [ \frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}\frac{a}{b} ]+\tan [ \frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}\frac{a}{b} ]= $
[MP PET 1999]
Options:
A) $ \frac{2a}{b} $
B) $ \frac{2b}{a} $
C) $ \frac{a}{b} $
D) $ \frac{b}{a} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \tan [ \frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}\frac{a}{b} ]+\tan [ \frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}\frac{a}{b} ] $
Let $ \frac{1}{2}{{\cos }^{-1}}\frac{a}{b}=\theta \Rightarrow \cos 2\theta =\frac{a}{b} $
Thus, $ \tan [ \frac{\pi }{4}+\theta ]+\tan [ \frac{\pi }{4}-\theta ] $ = $ \frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }=\frac{{{(1+\tan \theta )}^{2}}+{{(1-\tan \theta )}^{2}}}{(1-{{\tan }^{2}}\theta )} $
= $ \frac{1+{{\tan }^{2}}\theta +2\tan \theta +1+{{\tan }^{2}}\theta -2\tan \theta }{(1+{{\tan }^{2}}\theta )} $
= $ \frac{2(1+{{\tan }^{2}}\theta )}{1-{{\tan }^{2}}\theta }=2\sec 2\theta =\frac{2}{\cos 2\theta } $
= $ \frac{2}{a/b}=\frac{2b}{a} $ .
BETA
