Question

The value of expression : ${\tan }^{-1}\left(\frac{\sqrt{2}}{2}\right)+{\sin }^{-1}\left(\frac{\sqrt{5}}{5}\right)-{\cos }^{-1}\left(\frac{\sqrt{10}}{10}\right)$ is

• A. ${\cot }^{-1}\left(\frac{1+\sqrt{2}}{1-\sqrt{2}}\right)$
• B. ${\cot }^{-1}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$
• C. $-\pi +{\cot }^{-1}\left(\frac{1+\sqrt{2}}{1-\sqrt{2}}\right)$
• D. $\pi -{\cot }^{-1}\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}\right)$

Answer 1

Answer: (C)

${\tan }^{-1}\frac{1}{\sqrt{2}}+{\sin }^{-1}\frac{1}{\sqrt{5}}-{\cos }^{-1}\frac{1}{\sqrt{10}}\Rightarrow {\tan }^{-1}\frac{1}{\sqrt{2}}+{\tan }^{-1}\frac{1}{2}-{\tan }^{-1}3$
${\tan }^{-1}\frac{1}{\sqrt{2}}-\left[{\tan }^{-1}3-{\tan }^{-1}\frac{1}{2}\right]\Rightarrow {\tan }^{-1}\frac{1}{\sqrt{2}}-\left[{\tan }^{-1}\frac{3-\frac{1}{2}}{1+\frac{3}{2}}\right]=\left({\tan }^{-1}\frac{1}{\sqrt{2}}\right)\left({\tan }^{-1}1\right)$
$=-\left[{\tan }^{-1}1-{\tan }^{-1}\frac{1}{\sqrt{2}}\right]=-\left[{\tan }^{-1}\frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}\right]=-{\tan }^{-1}\frac{\sqrt{2}-1}{\sqrt{2}+1}$
$=-{\cot }^{-1}\frac{\sqrt{2}+1}{\sqrt{2}-1}=-\left[{\cot }^{-1}-\left(\frac{1+\sqrt{2}}{1-\sqrt{2}}\right)\right]=-\left[\pi -{\cot }^{-1}\frac{1+\sqrt{2}}{1-\sqrt{2}}\right]=-\pi +{\cot }^{-1}\frac{1+\sqrt{2}}{1-\sqrt{2}}\Rightarrow \text{ (C) }$