Question

If we have an expression as $u = \log \tan \left[ {\dfrac{\pi }{4} + \dfrac{x}{2}} \right]$ then, $\cosh u =$
1. $\sec x$
2. $\operatorname{cosec} x$
3. $\tan x$
4. $\sin x$

Answer 1

We know that the hyperbolic function $f\left( x \right) = \cosh x$ is defined as
$\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}$
$\Rightarrow \cosh u = \dfrac{{{e^u} + {e^{ - u}}}}{2}$
It is given that $u = \log \tan \left[ {\dfrac{\pi }{4} + \dfrac{x}{2}} \right]$
We know that according to the property of logarithm,
if ${\log _a}b = x$ then $b = {a^x}$
Therefore if $u = {\log _e}\tan \left[ {\dfrac{\pi }{4} + \dfrac{x}{2}} \right]$ then ${e^u} = \tan \left[ {\dfrac{\pi }{4} + \dfrac{x}{2}} \right]$
Substitute the value of ${e^u}$ in equation $\cosh u = \dfrac{{{e^u} + {e^{ - u}}}}{2}$ and then simplify the equation.

Complete step-by-step solution:
It is given that
$u = \log \tan \left[ {\dfrac{\pi }{4} + \dfrac{x}{2}} \right]$
$\Rightarrow {e^u} = \tan \left[ {\dfrac{\pi }{4} + \dfrac{x}{2}} \right]$
We are asked to find the value of $\cosh u$
We know that $\cosh u = \dfrac{{{e^u} + {e^{ - u}}}}{2}$
$\Rightarrow \cosh u = \dfrac{{{e^u}}}{2} + \dfrac{1}{{2{e^u}}}$
Taking L.C.M we get,
$\cosh u = \dfrac{{{e^{2u}} + 1}}{{2{e^u}}}$
Substituting the value of ${e^u}$ in above equation, we get
$\cosh u = \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right) + 1}}{{2\tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)}} - - - - - \left( 2 \right)$
We know that from trigonometric identity,
$\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$
Taking reciprocal we get,
$\dfrac{1}{{\sin 2x}} = \dfrac{{1 + {{\tan }^2}x}}{{2\tan x}}$
$\therefore \dfrac{1}{{\sin \left( {\dfrac{\pi }{2} + x} \right)}} = \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right) + 1}}{{2\tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)}}$
$\Rightarrow \cosh u = \dfrac{1}{{\sin \left( {\dfrac{\pi }{2} + x} \right)}}$
We know that $\sin \left( {\dfrac{\pi }{2} + x} \right) = \cos x$
$\therefore \cosh u = \dfrac{1}{{\cos x}}$
$\Rightarrow \cosh u = \sec x$ $\left( {\because \dfrac{1}{{\cos x}} = \sec x} \right)$
Hence the answer is $\sec x$

Note: This problem can be solved alternatively.
If $u = {\log _e}\tan \left[ {\dfrac{\pi }{4} + \dfrac{x}{2}} \right]$ then ${e^u} = \tan \left[ {\dfrac{\pi }{4} + \dfrac{x}{2}} \right]$
We know that $\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$
$\therefore {e^u} = \dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan \left( {\dfrac{x}{2}} \right)}}{{1 - \tan \left( {\dfrac{\pi }{4}} \right)\tan \left( {\dfrac{x}{2}} \right)}}$
$\Rightarrow {e^u} = \dfrac{{1 + \tan \left( {\dfrac{x}{2}} \right)}}{{1 - \tan \left( {\dfrac{x}{2}} \right)}}$ $\left( {\because \tan \left( {\dfrac{\pi }{4}} \right) = 1} \right)$
Substituting the value of ${e^u}$ in $\cosh u = \dfrac{{{e^{2u}} + 1}}{{2{e^u}}}$ , we get
$\cosh u = \dfrac{{{{\left( {\dfrac{{1 + \tan \left( {\dfrac{x}{2}} \right)}}{{1 - \tan \left( {\dfrac{x}{2}} \right)}}} \right)}^2} + 1}}{{2\left( {\dfrac{{1 + \tan \left( {\dfrac{x}{2}} \right)}}{{1 - \tan \left( {\dfrac{x}{2}} \right)}}} \right)}}$
Expanding and then simplifying we get
$\cosh u = \dfrac{1}{{\left( {\dfrac{{1 - {{\tan }^2}\left( {\dfrac{\pi }{2}} \right)}}{{1 + {{\tan }^2}\left( {\dfrac{\pi }{2}} \right)}}} \right)}}$
We know that $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$
$\therefore \cosh u = \dfrac{1}{{\cos x}}$
$\Rightarrow \cosh u = \sec x$