Question

The value of ${\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(3\right)$ is

• A. $\frac{3\mathrm{\pi }}{4}$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{3}$
• D. ${\tan }^{-1}\left(6\right)$
• E. ${\tan }^{-1}\left(5\right)$

Answer 1

The correct option is A

$\frac{3\mathrm{\pi }}{4}$

The explanation for the correct option

The given trigonometric expression: ${\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(3\right)$.

It is known that, ${\tan }^{-1}\left(A\right)+{\tan }^{-1}\left(B\right)={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$.

Thus, ${\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(3\right)={\tan }^{-1}\left(\frac{2+3}{1-\left(2\right)\left(3\right)}\right)$.

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

Hence, option A is correct .