Question

Solve for $x: \tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4} ; x>0$

Answer 1

SOLUTION : We have, $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4} \quad \Rightarrow \tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \cdot 3 x}\right)=\frac{\pi}{4}\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right]$

$\frac{5 x}{1-6 x^2}=\tan \frac{\pi}{4}  \Rightarrow  \frac{5 x}{1-6 x^2}=1 \\$

$5 x=1-6 x^2  \Rightarrow  6 x^2+6 x-x-1=0 \\$

$6 x^2+6 x-x-1=0  \Rightarrow  6 x(x+1)-1(x+1)=0 \\$

$x=-1, \text { or } x=\frac{1}{6} $

Since $x=-1$ does not satisfy the equation as the L.H.S. of equation becomes negative.

Hence $x=\frac{1}{6}$ is the required solution.