Question

If $ta{n}^{-1}4x+ta{n}^{-1}6x=\pi /4,$ 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)

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