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# The value of the determinant $\left| {\begin{array}{*{20}{c}}{{{\sin }^2}36^\circ }&{{{\cos }^2}36^\circ }&{\cot 135^\circ }\\{{{\sin }^2}53^\circ }&{\cot 135^\circ }&{{{\cos }^2}53^\circ }\\{\cot 135^\circ }&{{{\cos }^2}25^\circ }&{{{\cos }^2}65^\circ }\end{array}} \right|$ is A. $-2$B. $-1$C. 0D. 1E. 2

Last updated date: 13th Jun 2024
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Hint: Here we need to find the value of the determinant of the given matrix. The given matrix is of the order 3. So we will use the rules to find the determinant. Then we will use the basic trigonometric identities and then we will simplify each element of the matrix first.

Complete step by step solution:
We will first calculate the value of the given determinant i.e.
$D = \left| {\begin{array}{*{20}{c}}{{{\sin }^2}36^\circ }&{{{\cos }^2}36^\circ }&{\cot 135^\circ }\\{{{\sin }^2}53^\circ }&{\cot 135^\circ }&{{{\cos }^2}53^\circ }\\{\cot 135^\circ }&{{{\cos }^2}25^\circ }&{{{\cos }^2}65^\circ }\end{array}} \right|$
Here, we will use the basic trigonometric identities and then we will simplify each element of the matrix first.
We know that from the periodic trigonometric identities that $\sin \theta = \cos \left( {90^\circ - \theta } \right)$
Applying this identity, we get
${{\sin }^{2}}65{}^\circ =\cos \left( 90{}^\circ -65{}^\circ \right)={{\cos }^{2}}25{}^\circ$
Therefore, we get
$\Rightarrow D = \left| {\begin{array}{*{20}{c}}{{{\sin }^2}36^\circ }&{{{\cos }^2}36^\circ }&{\cot 135^\circ }\\{{{\sin }^2}53^\circ }&{\cot 135^\circ }&{{{\cos }^2}53^\circ }\\{\cot 135^\circ }&{{{\sin }^2}65^\circ }&{{{\cos }^2}65^\circ }\end{array}} \right|$
Now, we can write $\cot 135^\circ$as $\cot \left( {90^\circ + 45^\circ } \right)$ and we know the periodic trigonometric identities that $- \tan \theta = \cos \left( {90^\circ + \theta } \right)$
Applying this here, we get
$\cot \left( {90^\circ + 45^\circ } \right) = - \tan 45^\circ = - 1$
Substituting this value, we get
$\Rightarrow D = \left| {\begin{array}{*{20}{c}}{{{\sin }^2}36^\circ }&{{{\cos }^2}36^\circ }&{ - 1}\\{{{\sin }^2}53^\circ }&{ - 1}&{{{\cos }^2}53^\circ }\\{ - 1}&{{{\sin }^2}65^\circ }&{{{\cos }^2}65^\circ }\end{array}} \right|$
Now, we will apply, ${C_1} \to {C_1} + {C_2} + {C_3}$.
$\Rightarrow D = \left| {\begin{array}{*{20}{c}}{{{\sin }^2}36^\circ + {{\cos }^2}36^\circ - 1}&{{{\cos }^2}36^\circ }&{ - 1}\\{{{\sin }^2}53^\circ - 1 + {{\cos }^2}53^\circ }&{ - 1}&{{{\cos }^2}53^\circ }\\{ - 1 + {{\sin }^2}65^\circ + {{\cos }^2}65^\circ }&{{{\sin }^2}65^\circ }&{{{\cos }^2}65^\circ }\end{array}} \right|$
We know the trigonometric identity that ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Using this identity here, we get
$\Rightarrow D = \left| {\begin{array}{*{20}{c}}{1 - 1}&{{{\cos }^2}36^\circ }&{ - 1}\\{1 - 1}&{ - 1}&{{{\cos }^2}53^\circ }\\{1 - 1}&{{{\sin }^2}65^\circ }&{{{\cos }^2}65^\circ }\end{array}} \right|$
On further simplification, we get
$\Rightarrow D = \left| {\begin{array}{*{20}{c}}0&{{{\cos }^2}36^\circ }&{ - 1}\\0&{ - 1}&{{{\cos }^2}53^\circ }\\0&{{{\sin }^2}65^\circ }&{{{\cos }^2}65^\circ }\end{array}} \right|$
We know that if all the elements of any one of the rows or columns are zero then the value of the determinant is equal to zero.
Therefore, we get the value of determinant equal to zero.

Hence, the correct answer is option C.

Note:
Here we have obtained the value of the determinant of the matrix which is of order 3. So we need to know some basic properties of determinants. The most important property of a determinant is that if two or more rows are similar then the value of the determinant is equal to zero. Also, keep in mind that if we interchange the rows and columns the value of the determinant remains the same. If all the elements of any one of the rows or columns are zero then the value of the determinant is equal to zero. If we multiply all the elements of a row by a constant then the resultant determinant will be constant times the value of the original determinant.