Question

The value of the 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

The correct option is C $-\pi +{\cot }^{-1}\left(\frac{1+\sqrt{2}}{1-\sqrt{2}}\right)$

${\tan }^{-1}\left(\frac{\sqrt{2}}{2}\right)+{\sin }^{-1}\left(\frac{\sqrt{5}}{5}\right)-{\cos }^{-1}\left(\frac{\sqrt{10}}{10}\right)$
$={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)+{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)-{\cos }^{-1}\left(\frac{1}{\sqrt{10}}\right)$
$={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}3$
$={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)-[{\tan }^{-1}3-{\tan }^{-1}\left(\frac{1}{2}\right)]$
$={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)-{\tan }^{-1}\left(\frac{3-\frac{1}{2}}{1+\frac{3}{2}}\right)$
$={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)-{\tan }^{-1}1$
$=-[{\tan }^{-1}1-{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)]$
$=-\left[{\tan }^{-1}\left(\frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}\right)\right]$
$=-{\tan }^{-1}\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)$
$=-{\cot }^{-1}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)[\because {\tan }^{-1}x={\cot }^{-1}\left(\frac{1}{x}\right)\text{for}x>0]$
$=-{\cot }^{-1}(-\left(\frac{1+\sqrt{2}}{1-\sqrt{2}}\right))$
$=-[\pi -{\cot }^{-1}\left(\frac{1+\sqrt{2}}{1-\sqrt{2}}\right)][\because {\cot }^{-1}(-x)=\pi -{\cot }^{-1}x\forall x\in R]$
$=-\pi +{\cot }^{-1}\left(\frac{1+\sqrt{2}}{1-\sqrt{2}}\right)$