Question

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

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

Answer 1

The correct option is A

$secx$

Explanation for the correct answer:

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

Since no base of logarithm is given we consider natural logarithm

$\Rightarrow$$u={\log }_e\left(\frac{1+\tan {\displaystyle \frac{x}{2}}}{1-\tan {\displaystyle \frac{x}{2}}}\right)$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \tan \left(A+B\right)=\frac{\mathrm{tanA}+\mathrm{tanB}}{1-\mathrm{tanAtanB}},\tan \frac{\pi }{4}=1\right]$

Taking exponential on both sides we get

$\Rightarrow$$e^u=e^{\log \left(\frac{1+\tan {\displaystyle \frac{x}{2}}}{1-\tan {\displaystyle \frac{x}{2}}}\right)}$

$\Rightarrow$$e^u={\left(\frac{1+\tan {\displaystyle \frac{x}{2}}}{1-\tan {\displaystyle \frac{x}{2}}}\right)}^{\log e}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because a^{\log b}=b^{\log a},{\log }_{e}e=1\right]$

$\Rightarrow$$e^u=\frac{1+\tan {\displaystyle \frac{x}{2}}}{1-\tan {\displaystyle \frac{x}{2}}}$ $...\left(i\right)$

$\Rightarrow$$e^{-u}=\frac{1-\tan {\displaystyle \frac{x}{2}}}{1+\tan {\displaystyle \frac{x}{2}}}$ $...\left(ii\right)$

Hyperbolic cosine can be written as

$\cos hu=\frac{e^u+e^{-u}}{2}$

Substituting the values from we get

$\cos hu=\frac{\frac{1+\tan {\displaystyle \frac{x}{2}}}{1-\tan {\displaystyle \frac{x}{2}}}+\frac{1-\tan {\displaystyle \frac{x}{2}}}{1+\tan {\displaystyle \frac{x}{2}}}}{2}$

$\Rightarrow$$2\cos hu=\frac{1+\tan {\displaystyle \frac{x}{2}}}{1-\tan {\displaystyle \frac{x}{2}}}+\frac{1-\tan {\displaystyle \frac{x}{2}}}{1+\tan {\displaystyle \frac{x}{2}}}$

$\Rightarrow$$2\cos hu=\frac{{\displaystyle {\left(1+\tan \frac{x}{2}\right)}^2+{\left(1-\tan \frac{x}{2}\right)}^2}}{{\displaystyle \left(1+\tan \frac{{\displaystyle x}}{{\displaystyle 2}}\right)\left(1-\tan \frac{x}{2}\right)}}$

$\Rightarrow$$2\cos hu=\frac{{\displaystyle 2\left(1+{\tan }^2\frac{x}{2}\right)}}{{\displaystyle 1-{\tan }^2\frac{x}{2}}}$

$\Rightarrow$ $\cos hu=\frac{{\displaystyle \left(\frac{{\sin }^2{\displaystyle \frac{x}{2}}+{\cos }^2{\displaystyle \frac{x}{2}}}{{\cos }^2{\displaystyle \frac{x}{2}}}\right)}}{{\displaystyle \left(\frac{{\displaystyle {\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}}{{\cos }^2{\displaystyle \frac{x}{2}}}\right)}}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \tan \frac{x}{2}=\frac{\sin {\displaystyle \frac{x}{2}}}{\cos {\displaystyle \frac{x}{2}}}\right]$

$\Rightarrow$ $\cos hu=\frac{{\displaystyle 1}}{{\displaystyle 2{\cos }^2\frac{{\displaystyle x}}{{\displaystyle 2}}-1}}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}=1\right]$

$\Rightarrow$ $\cos hu=\frac{{\displaystyle 1}}{{\displaystyle \cos x}}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \mathrm{cosx}=2{\cos }^2\frac{x}{2}-1\right]$

$\Rightarrow$ $\cos hu=secx$

Hence the value of $\cos hu$ is $secx$ for the given value of $u$,

Hence, option(A) is the correct answer