Question

lf coshx $=\sec \Theta$, then ${\coth }^2\left(x/2\right)=$

• A. ${\tan }^2\left(\Theta /2\right)$
• B. ${\tan }^2\Theta$
• C. ${\cot }^2\left(\Theta /2\right)$
• D. ${\cot }^2\Theta$

Answer 1

Answer: (C)

${\cot }^2\left(\Theta /2\right)$

Given $coshx=sec\theta =2cos{h}^2\frac{x}{2}-1$
we know that,
$a+ta{n}^2\theta =se{c}^2\theta$
$1+co{t}^2\theta =cose{c}^2\theta$
$1+ta{n}^2\theta =cos{h}^{2}x$
$ta{n}^2\theta =si{n}^{2}hx$
$\therefore cot{h}^2\frac{x}{2}=\frac{1}{tan{h}^2\frac{x}{2}}$
$\frac{1}{tan{h}^2\frac{x}{2}}=\frac{1}{1-\frac{2tanh\frac{x}{2}}{tanhx}}[tan2x=\frac{2tanx}{1-ta{n}^{2}x}]$
$\frac{1}{tan{h}^2\frac{x}{2}}=\frac{tanhx}{tanhx-2tanh\frac{x}{2}}$
$tanhx-2tanh\frac{x}{2}=tanhxtan{h}^2\frac{x}{2}$