Inverse Trigonometric Functions Question 218
Question: If $ {{\tan }^{-1}}\frac{x-1}{x+1}+{{\tan }^{-1}}\frac{2x-1}{2x+1}={{\tan }^{-1}}\frac{23}{36}, $ then x =
[ISM Dhanbad 1973]
Options:
A) $ \frac{3}{4},\frac{-3}{8} $
B) $ \frac{3}{4},\frac{3}{8} $
C) $ \frac{4}{3},\frac{3}{8} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{x-1}{x+1}+\frac{2x-1}{2x+1}=\frac{23}{36}[ 1-\frac{(x-1)(2x-1)}{(x+1)(2x+1)} ] $
Therefore $ 24x^{2}-23x-12=0\Rightarrow (3x-4)(8x+3)=0 $ . Hence, the solutions are $ x = \frac{4}{3} $ and $ x = -\frac{3}{8} $.
BETA
