${\tan ^{ - 1}}\sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} , - \frac{\pi }{4} < x < \frac{\pi }{4}$

Method I

Let $y = {\tan ^{ - 1}}\sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}}$

therefore,$\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\tan ^{ - 1}}\sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}}$

$= \frac{1}{{1 + \sqrt {{{\left( {\frac{{1 - \cos x}}{{1 + \cos x}}} \right)}^2}} }} \cdot \frac{d}{{dx}}{\left[ {\frac{{1 - \cos x}}{{1 + \cos x}}} \right]^{1/2}}$

$= \frac{1}{{1 + \frac{{1 - \cos x}}{{1 + \cos x}}}} \cdot \frac{1}{2}{\left[ {\frac{{1 - \cos x}}{{1 + \cos x}}} \right]^{ - 1/2}} \cdot \frac{d}{{dx}}\left( {\frac{{1 - \cos x}}{{1 + \cos x}}} \right)$

$= \frac{1}{{\frac{{1 + \cos x + 1 - \cos x}}{{1 + \cos x}}}} \cdot \frac{1}{2}{\left[ {\frac{{(1 - \cos x)}}{{(1 + \cos x)}} \cdot \frac{{(1 - \cos x)}}{{(1 - \cos x)}}} \right]^{ - 1/2}}$
$\cdot \frac{{(1 + \cos x) \cdot \sin x + (1 - \cos x) \cdot \sin x}}{{{{(1 + \cos x)}^2}}}$

$= \frac{{(1 + \cos x)}}{2} \cdot \frac{1}{2}{\left[ {\frac{{{{(1 - \cos x)}^2}}}{{\left( {1 - {{\cos }^2}x} \right)}}} \right]^{ - 1/2}}\left[ {\frac{{\sin x(1 + \cos x + 1 - \cos x)}}{{{{(1 + \cos x)}^2}}}} \right]$

$= \frac{{(1 + \cos x)}}{2} \cdot \frac{1}{2}{\left[ {\frac{{{{(1 - \cos x)}^2}}}{{\left( {1 - {{\cos }^2}x} \right)}}} \right]^{ - 1/2}}\left[ {\frac{{\sin x(1 + \cos x + 1 - \cos x)}}{{{{(1 + \cos x)}^2}}}} \right]$

$= \frac{{(1 + \cos x)}}{2} \cdot \frac{1}{2}{\left[ {\frac{{{{(1 - \cos x)}^2}}}{{\sin x}}} \right]^{ - 1/2}} \cdot \frac{{2\sin x}}{{{{(1 + \cos x)}^2}}}$

$= \frac{{(1 + \cos x)}}{2} \cdot \frac{1}{2} \cdot \frac{{\sin x}}{{(1 - \cos x)}} \cdot \frac{{2\sin x}}{{{{(1 + \cos x)}^2}}}$

$= \frac{{2{{\sin }^2}x}}{{4(1 + \cos x)(1 - \cos x)}} = \frac{1}{2} \cdot \frac{{{{\sin }^2}x}}{{\left( {1 - {{\cos }^2}x} \right)}}$

$= \frac{1}{2} \cdot \frac{{{{\sin }^2}x}}{{{{\sin }^2}x}} = \frac{1}{2}$

$\Rightarrow \frac{dy}{dx} = \frac{1}{2} \cdot \frac{{{{\sin }^2}x}}{{{{\sin }^2}x}} = \frac{1}{2}$

Method II

Let $y = {\tan ^{ - 1}}\left( {\sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} } \right)$

$= {\tan ^{ - 1}}\left( {\tan \frac{x}{2}} \right) = \frac{x}{2}$
On differentiating w.r.t. $x$, we get
$\frac{{dy}}{{dx}} = \frac{1}{2}$