If $y=\sqrt{\dfrac{1-\sin 2x}{1+\sin 2x}}$, then $\left(\dfrac{dy}{dx}\right)_{x=0}=$

$y=\sqrt { \dfrac { 1-\sin2x }{ 1+\sin2x } }$

$=\sqrt { \dfrac { 1-2sinxcosx }{ 1+2sinxcosx } }$

$=\sqrt { \dfrac { { \sin }^{ 2 }x+{ \cos }^{ 2 }x-2sinxcosx }{ { \sin }^{ 2 }x+{ \cos }^{ 2 }x+2sinxcosx } }$

$=\sqrt { { \left( \dfrac { sinx-cosx }{ sinx+cosx } \right) }^{ 2 } }$

$=\pm \dfrac { \left( sinx-cosx \right) }{ \left( sinx+cosx \right) } \times \dfrac { \left( sinx-cosx \right) }{ \left( sinx-cosx \right) }$

$=\pm \dfrac { { \sin }^{ 2 }x+{ \cos }^{ 2 }x-2sinxcosx }{ { \sin }^{ 2 }x-{ \cos }^{ 2 }x }$

$=\pm \dfrac { 1-\sin2x }{ \cos2x }$

$y=\mp \sec2x-\tan2x$

$\dfrac { dy }{ dx } =\pm \left( 2\sec2x\tan2x-2{ \sec }^{ 2 }2x \right)$

At $x=0$

$\dfrac { dy }{ dx } =\pm \left( -2 \right) \quad \Rightarrow \pm 2$

Answer (D)