Inverse Trigonometric Functions Question 202
Question: If $ {{\tan }^{-1}}2x+{{\tan }^{-1}}3x=\frac{\pi }{4} $ , then x =
[Roorkee 1978, 80; MNR 1986; Pb. CET 2001; Karnataka CET 2002]
Options:
A) - 1
B) $ \frac{1}{6} $
C) $ -1,\frac{1}{6} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{\tan }^{-1}}2x+{{\tan }^{-1}}3x=\frac{\pi }{4} $
$ \Rightarrow {{\tan }^{-1}}( \frac{2x+3x}{1-(2x)(3x)} )=\frac{\pi }{4} $
$ \Rightarrow {{\tan }^{-1}}( \frac{5x}{1-6x^{2}} )={{\tan }^{-1}}(1) $
$ \Rightarrow \frac{5x}{1-6x^{2}}=1 $
$ \Rightarrow 1-6x^{2}=5x $
$ \Rightarrow 6x^{2}+5x-1=0 $
$ \Rightarrow (x+1)( x-\frac{1}{6} )=0 $
$ \Rightarrow x=-1,\frac{1}{6} $ But - 1 does not hold. Trick: Check with the options. Obviously the equation holds for $ x=\frac{1}{6} $ , but not for - 1.
BETA
