Question

lf $y=\frac{1}{2}$ cosech-1 $(\frac{1}{2_X\sqrt{1+x^2}})$ , then $x=$

• A. $\cosh y$
• B. $\sinh y$
• C. $\tanh y$
• D. $\coth y$

Answer 1

Answer: (B)

$\sinh y$

Given $y=\frac{1}{2}cosec{h}^{-1}\left(\frac{1}{2x\sqrt{1+x^2}}\right)=\frac{1}{2}ln\left(\frac{1+\sqrt{1-\left(\frac{1}{4x^2(1+x^2)}\right)}}{\frac{1}{2x\sqrt{1+x^2}}}\right)$
$=\frac{1}{2}ln\left(\frac{\sqrt{4x^2+4x^4-1}+2x\sqrt{1+x^2}}{1}\right)$
$e^{2y}=\sqrt{4x^2+4x^4-1}+2x\sqrt{2+x^2}$
$e^{2y}=\frac{1}{2x\sqrt{1+x^2}-\sqrt{4x^4+4x^2-1}}$
$2x\sqrt{1+x^2}=\frac{e^{2y}+e^{-2y}}{2}=cosh2y$
and $\sqrt{4x^4+4x^2-1}=\frac{e^{2y}-e^{-2y}}{2}=sinh2y$
$\therefore 4x^4+4x^2-(1+si{n}^{2}h2y)=0$
$4x^4+4x^2+co{s}^{2}hy=0$
$x^2=\frac{-4\pm \sqrt{16-16co{s}^{2}hy}}{8}$
$x^2=\frac{-1\pm isinhy}{2}$