Inverse Trigonometric Functions Question 185
Question: $ 2{{\tan }^{-1}}\frac{1}{3}+{{\tan }^{-1}}\frac{1}{2}= $
Options:
A) $ 90^{o} $
B) $ 60^{o} $
C) $ 45^{o} $
D) $ {{\tan }^{-1}}2 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ 2{{\tan }^{-1}}\frac{1}{3}+{{\tan }^{-1}}\frac{1}{2}={{\tan }^{-1}}( \frac{\frac{2}{3}}{1-\frac{1}{9}} )+{{\tan }^{-1}}( \frac{1}{2} ) $
$ ={{\tan }^{-1}}( \frac{\frac{2}{3}}{\frac{8}{9}} )+{{\tan }^{-1}}( \frac{1}{2} )={{\tan }^{-1}}( \frac{3}{4} )+{{\tan }^{-1}}( \frac{1}{2} ) $
$ ={{\tan }^{-1}}( \frac{\frac{1}{2}+\frac{3}{4}}{1-\frac{1}{2}\times \frac{3}{4}} )={{\tan }^{-1}}(2) $ .
BETA
