Question

lf $x=\log$ Tan $(\frac{\pi }{4}+\frac{\theta }{2})$ ,then $\cos h\left(x\right)=$

• A. $\cos \theta$
• B. $\sinh \theta$
• C. $\sec \theta$
• D. $\left[cosec\right]\theta$

Answer 1

Answer: (C)

$\sec \theta$

we know by identity,
$coshx=\frac{e^x+e^{-x}}{2}$
Given $x=logtan(\frac{\pi }{4}+\frac{\theta }{2})$
$\Rightarrow coshx=\frac{e^{logtan(\frac{\pi }{4}+\frac{\theta }{2})}+e^{-logtan(\frac{\pi }{4}+\frac{\theta }{2})}}{2}$
$=\frac{tan(\frac{\pi }{4}+\frac{\theta }{2})+cot(\frac{\pi }{4}+\frac{\theta }{2})}{2}$
$=\frac{1}{2sin(\frac{\pi }{4}+\frac{\theta }{2})cos(\frac{\pi }{4}+\frac{\theta }{2})}$
$=\frac{1}{sin(\frac{\pi }{2}+\theta )}[2sinxcosx=sin2x]$
$\Rightarrow coshx=\frac{1}{cos\theta }=sec\theta$