Question

If $ta{n}^{-1}2x+ta{n}^{-1}3x=\frac{\pi }{4}$, then $x$ is

• A. $\frac{1}{6}$
• B. $-1$
• C. $\left(\frac{1}{6},-1\right)$
• D. None of these

Answer 1

Answer: (A)

$ta{n}^{-1}2x+ta{n}^{-1}3x=\frac{\pi }{4}$
$\Rightarrow \frac{3x+2x}{1-6x^2}=tan\frac{\pi }{4}$
$\Rightarrow 5x=1-6x^2$
$\Rightarrow 6x^2+5x-1=0$
$\Rightarrow x=-1,\frac{1}{6}$
But when $x=-1$, $ta{n}^{-1}2x=ta{n}^{-1}\left(-2\right)<0$ and $ta{n}^{-1}3x=ta{n}^{-1}\left(-3\right)<0$
This value will not satisfy the given equation.
Hence, $x=\frac{1}{6}$.