Question

If $ tan^{-1} 4x + tan^{-1} 6x = \pi /4, $ then $ x $ is equal to

• A. $ \frac {1}{12} $
• B. $ \frac {1}{12} or - \frac{1}{2} $
• C. $ - \frac {1}{2} $
• D. None of these

Answer 1

Answer: (B)

$tan^{-1}\left(4x\right)+tan^{-1}\left(6x\right)=\pi/ 4$
$\Rightarrow tan^{-1} \left(\frac{4x+6x}{1-24x^{2}}\right)=\pi /4 $
$\Rightarrow \frac{10x}{1-24x^{2}} =tan\left(\pi /4\right)=1 $
$\Rightarrow x\cdot10=1-24x^{2}$
$\Rightarrow 24x^{2}+10x-1=0$
$\Rightarrow 24x^{2}+12x-2x-1=0$
$\Rightarrow 12x\left(2x+1\right)-1\left(2x+1\right)=0$
$\Rightarrow \left(2x+1\right)\left(12x-1\right)=0$
$\Rightarrow x=-1 /2$ or $1 /12$