Question

lf '$x$' is an acute angle and $y=\log \left(\tan \right(\frac{\pi }{4}+\frac{x}{2}\left)\right)$ then cosx. coshy $=\dots \dots$..

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Given $y=log\left(tan\right(\frac{\pi }{4}+\frac{x}{2}\left)\right),x$ is acute angle
$coshx=\frac{e^x+e^{-x}}{2}$ [by identity]
$\therefore coshy=\frac{e^y+e^{-y}}{2}=\frac{tan(\frac{\pi }{4}+\frac{x}{2})+cot(\frac{\pi }{4}+\frac{x}{2})}{2}$
$=\frac{1}{sin2(\frac{\pi }{4}+\frac{x}{2})}[si{n}^2\theta +co{s}^2\theta =1,2sinxcosx=sin2x]$
$coshy=\frac{1}{cosx}=secx$
$\therefore \Rightarrow coshy.cosx=1.$