Inverse Trigonometric Functions Question 79
Question: $ 2{{\tan }^{-1}}( \frac{1}{3} )+{{\tan }^{-1}}( \frac{1}{7} )= $
[EAMCET 1983]
Options:
A) $ {{\tan }^{-1}}( \frac{49}{29} ) $
B) $ \frac{\pi }{2} $
C) 0
D) $ \frac{\pi }{4} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ 2{{\tan }^{-1}}( \frac{1}{3} )+{{\tan }^{-1}}( \frac{1}{7} )={{\tan }^{-1}}( \frac{2(1/3)}{1-(1/9)} )+{{\tan }^{-1}}( \frac{1}{7} ) $
$ ={{\tan }^{-1}}\frac{3}{4}+{{\tan }^{-1}}\frac{1}{7}={{\tan }^{-1}}( \frac{(3/4)+(1/7)}{1-(3/4)\times (1/7)} ) $
$ ={{\tan }^{-1}}( \frac{25}{25} )=\frac{\pi }{4} $ .
BETA
