Q65 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics

$2 \operatorname{coth} h^{-1}(4)+\sec h^{-1}\left(\frac{3}{5}\right)=$

  • Correct answer
    A
    $\log 5$
  • B
    $2 \log 3$
  • C
    $3 \log 2$
  • D
    $\log \frac{5}{3}$

Solution

As $\cot h^{-1}(x)=\frac{1}{2} \log \left(\frac{x+1}{x-1}\right)$
and $\sec h^{-1}(x)=\log \left(\frac{1+\sqrt{1-x^2}}{x}\right)$
Now, $2 \cot h^{-1}(4)+\sec h^{-1}\left(\frac{3}{5}\right)$
$=\log \left(\frac{4+1}{4-1}\right)+\log \left(\frac{1+\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}\right)=\log \frac{5}{3}+\log 3=\log 5$

Hence, the correct answer is option: 'A'.

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