Inverse Trigonometric Functions Question 233
Question: $ \theta ={{\tan }^{-1}}(2tan^{2}\theta )-ta{n^{-1}}( \frac{1}{3}\tan \theta ) $ then $ \tan \theta = $
Options:
A) $ -2 $
B) $ -1 $
C) $ 2/3 $
D) $ 2 $
Show Answer
Answer:
Correct Answer: A
$ \theta ={{\tan }^{-1}}[ \frac{2{{\tan }^{2}}\theta -\frac{1}{3}\tan \theta }{1+\frac{2}{3}{{\tan }^{3}}\theta } ] $
$ \Rightarrow \tan \theta =\frac{6{{\tan }^{2}}\theta -\tan \theta }{3+2{{\tan }^{3}}\theta }\Rightarrow 1=\frac{6\tan \theta -1}{3+2{{\tan }^{3}}\theta } $
Or $ \tan \theta =0 $
$ \Rightarrow 2{{\tan }^{3}}\theta -6\tan \theta +4=0 $
$ \Rightarrow {{(tan\theta -1)}^{2}}(tan\theta +2)=0 $
$ \Rightarrow \tan \theta =1;\tan \theta =-2;\tan \theta =0. $
BETA
