Q65 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

$\operatorname{sech}^2\left(\tanh ^{-1} \frac{1}{2}\right)+\operatorname{cosech}^2\left(\operatorname{coth}^{-1} 3\right)=$

  • A
    $\frac{35}{9}$
  • B
    $\frac{3}{2}$
  • C
    $\frac{25}{4}$
  • Correct answer
    D
    $\frac{35}{4}$

Solution

$\begin{aligned} & \text { } \operatorname{sech}^2\left(\tanh ^{-1} \frac{1}{2}\right)+\operatorname{cosech}^2\left(\cot h^{-1}(3)\right) \\ & =\operatorname{sech}^2\left(\operatorname{sech}^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)+\operatorname{cosech}^2\left(\operatorname{cosech}^{-1} \sqrt{8}\right) \\ & =\frac{3}{4}+8=\frac{35}{4}\end{aligned}$

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

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