Question

$\cos \left[{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right]=?$

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{2}$
• D. $\frac{\mathrm{\pi }}{4}$

Answer 1

The correct option is A

$\frac{1}{\sqrt{2}}$

Explanation for the correct option:

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

Use the identity $\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{y}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{(}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{y}\mathbf{)}}{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{–}\mathbf{}\mathbf{x}\mathbf{y}\mathbf{)}}\mathbf{;}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{}\mathit{x}\mathit{y}\mathbf{}\mathbf{<}\mathbf{}\mathbf{1}$

$=ta{n}^{-1}\left[\frac{{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{2}}}{1–\left({\displaystyle \frac{1}{3}}\right)\left({\displaystyle \frac{1}{2}}\right)}\right]$

$={\tan }^{-1}\left[\frac{(2+3)}{(6–1)}\right]$

$={\tan }^{-1}\left(\frac{5}{5}\right)$

$={\tan }^{-1}\left(1\right)$

$={\tan }^{-1}\left[\tan \left(\frac{\pi }{4}\right)\right]$ ; $\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{=}\mathbf{1}\right]$

$=\frac{\pi }{4}$

$\therefore \cos \left[{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right]=\cos \frac{\pi }{4}$

$=\frac{\mathbf{1}}{\sqrt{2}}$

Hence, Option ‘A’ is Correct.