Answer
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429.3k+ views
Hint: This question can be solved by simply solving the determinant.
Given determinant is
$\left| {\begin{array}{*{20}{c}}
{\cos {{15}^ \circ }}&{\sin {{15}^ \circ }} \\
{\sin {{75}^ \circ }}&{\cos {{75}^ \circ }}
\end{array}} \right|$
Now on solving the determinant we get,
$\cos {75^ \circ } \cdot \cos {15^ \circ } - \sin {75^ \circ } \cdot \sin {15^ \circ }$
Now we know that,
$\cos \left( {A + B} \right) = \cos A \cdot \cos B - \sin A \cdot \sin B$
Using the above equation we get,
$
\cos \left( {{{75}^ \circ } + {{15}^ \circ }} \right) \\
{\text{or }}\cos \left( {{{90}^ \circ }} \right) \\
= 0 \\
$
Therefore, the correct option is (B).
Note: These types of questions can be solved by simply solving the determinant. Here in this question we simply solve the determinant and then we apply the formula of $\cos \left( {A + B} \right)$ and then we get our answer.
Given determinant is
$\left| {\begin{array}{*{20}{c}}
{\cos {{15}^ \circ }}&{\sin {{15}^ \circ }} \\
{\sin {{75}^ \circ }}&{\cos {{75}^ \circ }}
\end{array}} \right|$
Now on solving the determinant we get,
$\cos {75^ \circ } \cdot \cos {15^ \circ } - \sin {75^ \circ } \cdot \sin {15^ \circ }$
Now we know that,
$\cos \left( {A + B} \right) = \cos A \cdot \cos B - \sin A \cdot \sin B$
Using the above equation we get,
$
\cos \left( {{{75}^ \circ } + {{15}^ \circ }} \right) \\
{\text{or }}\cos \left( {{{90}^ \circ }} \right) \\
= 0 \\
$
Therefore, the correct option is (B).
Note: These types of questions can be solved by simply solving the determinant. Here in this question we simply solve the determinant and then we apply the formula of $\cos \left( {A + B} \right)$ and then we get our answer.
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