Question

If $u=\log \tan \left[\left(\frac{\mathrm{\pi }}{4}\right)+\left(\frac{x}{2}\right)\right]$, then $\cos hu=$

• A. $secx$
• B. $\cos ecx$
• C. $\tan x$
• D. $\sin x$

Answer 1

The correct option is A

$secx$

Explanation for the correct option:

Find the value of $\cos hu$ :

Given,

$$\begin{gathered} u=\log \tan \left[\left(\frac{\mathrm{\pi }}{4}\right)+\left(\frac{x}{2}\right)\right] \\ \Rightarrow e^u=\tan \left[\left(\frac{\mathrm{\pi }}{4}\right)+\left(\frac{x}{2}\right)\right] \\ \Rightarrow e^u=\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)} \end{gathered}$$

We know,

$\cos hu=\frac{\left[e^u+e^{-u}\right]}{2}=\frac{e^{2u}+1}{2e^u}$

Then,

$\begin{array}{rcl}\cos hu& =& \frac{{\left\{\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right\}}^2+1}{2\left[\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right]}\\ & =& \frac{{\left[1+\tan \left(\frac{x}{2}\right)\right]}^2+{\left[1-\tan \left(\frac{x}{2}\right)\right]}^2}{2\left[1+\tan \left(\frac{x}{2}\right)\right]\left[1-\tan \left(\frac{x}{2}\right)\right]}\\ & =& \frac{2+2{\tan }^2\left(\frac{x}{2}\right)}{2\left[1-{\tan }^2\left(\frac{x}{2}\right)\right]}\\ & & \\ & =& \frac{1+{\tan }^2\left(\frac{x}{2}\right)}{1-{\tan }^2\left(\frac{x}{2}\right)}\\ & =& \frac{1}{\cos x}\left[\because \cos 2x=\frac{1+{\tan }^{2}x}{1-{\tan }^{2}x}\right]\\ & =& secx\end{array}$

Hence, the correct option is A.