Mathematics | TS EAMCET 2018 | Discussion Page

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

$\log \left(\frac{4+2\sqrt{3}+2\sqrt{5}+\sqrt{15}}{\sqrt{15}}\right)$

$\log \left(\frac{4+\sqrt{3}+\sqrt{5}+\sqrt{15}}{\sqrt{15}}\right)$

$\log \left(\frac{(2+\sqrt{3})+(2+\sqrt{5})\sqrt{5}}{\sqrt{3}}\right)$

$\log \frac{(2+\sqrt{3})+(2+\sqrt{5})\sqrt{3}}{\sqrt{5}}$

Option D

We have

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

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

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

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

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

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

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