Question

lf $x=\log \left[\cot \right(\frac{\pi }{4}+\theta \left)\right]$ then sinh(x) $=\dots \dots$

• A. $\tan \left(2\theta \right)$
• B. $Cot2\theta$
• C. $-Tan2\theta$
• D. $-Cot2\theta$

Answer 1

The correct option is C $-Tan2\theta$

We know by identity of hyperbolic functions,
$sinhx=\frac{e^x-e^{-1}}{2}$
Given, $x=logcot(\frac{\pi }{4}+\theta )$
$\therefore sinhx=\frac{e^{logcot(\frac{\pi }{4}+\theta )}-e^{-logcot(\frac{\pi }{4}+\theta )}}{2}$
$=\frac{cot(\pi /4+\theta )-tan(\pi /4+\theta )}{2}$
$=\frac{co{s}^2(\pi /4+\theta )-si{n}^2(\pi /4+\theta )}{2sin(\pi /4+\theta )cos(\pi /4+\theta )}$
$=\frac{cos(\pi /2+2\theta )}{sin(\pi /2+2\theta )}[cos2x=co{s}^{2}x-si{n}^{2}x,sin2x=2sinxcosx]$
$=cot(\pi /2+2\theta )$
$\Rightarrow sinhx=-tan2\theta$