(iii) As per the question it is given that,

$\left| \begin{matrix}

\cos {{15}^{\circ }} & \sin {{15}^{\circ }}  \\

\sin {{75}^{\circ }} & \cos {{75}^{\circ }}  \\

\end{matrix} \right|$

$\Rightarrow \left| A \right|=\cos {{15}^{\circ }}\times \cos {{75}^{\circ }}+\sin {{15}^{\circ }}\times \sin {{75}^{\circ }}$

We all know the trigonometric function $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$

By putting this we have, $\left| A \right|=\cos {{\left( 75-15 \right)}^{\circ }}$

$\left| A \right|=\cos {{60}^{\circ }}$

$\left| A \right|=0.5$

(iv) As per the question it is given that,

$\left| \begin{matrix}

a+ib & c+id  \\

-c+id & a-ib  \\

\end{matrix} \right|$

$\Rightarrow \left| A \right|=\left( a+ib \right)\left( a-ib \right)-\left( c+id \right)\left( -c+id \right)$

$=\left( a+ib \right)\left( a-ib \right)+\left( c+id \right)\left( c-id \right)$

$={{a}^{2}}-{{i}^{2}}{{b}^{2}}+{{c}^{2}}-{{i}^{2}}{{d}^{2}}$

We all know that ${{i}^{2}}=-1$

$={{a}^{2}}-\left( -1 \right){{b}^{2}}+{{c}^{2}}-\left( -1 \right){{d}^{2}}$

$={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}$