Question

The number of solutions of the equation
${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\mathrm{\pi }}{4}$ is
(a) 2
(b) 3
(c) 1
(d) none of these

Answer 1

(a) 2
We know that ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$.

$$\begin{gathered} \therefore {\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{2x+3x}{1-2x\times 3x}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{2x+3x}{1-2x\times 3x}=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{5x}{1-6x^2}=1 \\ \Rightarrow 5x=1-6x^2 \\ \Rightarrow 6x^2+5x-1=0 \end{gathered}$$

Therefore, there are two solutions.