Q68 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

If $x=\log \left(y+\sqrt{y^2+1}\right)$ then $y=$

  • A
    $\tan h x$
  • B
    $\operatorname{cot h} \mathrm{x}$
  • Correct answer
    C
    $\sin h \mathrm{x}$
  • D
    $\cos h x$

Solution

$x=\log \left(y+\sqrt{\left.y^2+1\right)}\right.$ $\begin{aligned} & y+\sqrt{y^2+1}=e^x \\ & \because \sin h^{-1} x=\log \left(x+\sqrt{x^2+1}\right) \\ & \sin h^{-1} y=\log \left(y+\sqrt{y^2+1}\right) \\ & \sin h^{-1} y=x \Rightarrow \sin h x=y\end{aligned}$

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

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