Discussion Forum

$\sin h^{-1}2+\cos h^{-1}2-\tan h^{-1} \frac{2}{3}+\cot h^{-1}(-2)=$

Answer:

Explanation:

 We have 

$\sin h^{-1}2+\cos h^{-1}2-\tan h^{-1}\frac{ 2}{3}+\cot h^{-1}(-2)$

 = $ln(2+\sqrt{2^{2}+1})+ln(2+\sqrt{2^{2}-1})-\frac{1}{2}ln\left(\frac{1+2/3}{1-2/3}\right)+\frac{1}{2}ln\left(\frac{-2+1}{-2-1}\right)$

=$ln(2+\sqrt{5})+ln(2+\sqrt{3})-\frac{1}{2}ln 5+\frac{1}{2} ln \frac{1}{3}$

=$ln\left[\frac{(2+\sqrt{5})(2+\sqrt{3})\sqrt{3}}{\sqrt{5}}\right]$

                    $\because \sin h^{-1}x=ln(x+\sqrt{x^{2}+1)}, \cos h^{-1}x=ln(x+\sqrt{x^{2}-1)}$,

             $\tan h^{-1}x=\frac{1}{2}ln\left(\frac{1+x}{1-x}\right), \cot h^{-1}x=\frac{1}{2}ln \left(\frac{x+1}{x-1}\right)$