
Find the value of $\cos \theta \left[ {\begin{array}{*{20}{c}}
{\cos \theta }&{\sin \theta } \\
{ - \sin \theta }&{\cos \theta }
\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}
{\sin \theta }&{ - \cos \theta } \\
{\cos \theta }&{\sin \theta }
\end{array}} \right] = $
A $\left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$
B $\left[ {\begin{array}{*{20}{c}}
1&0 \\
0&0
\end{array}} \right]$
C $\left[ {\begin{array}{*{20}{c}}
0&1 \\
1&0
\end{array}} \right]$
D $\left[ {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right]$
Answer
160.8k+ views
Hint:First we will simplify the matrices then add the matrices. Then we will use the trigonometric identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$ to solve the question and get the required solution.
Formula Used: ${\cos ^2}\theta + {\sin ^2}\theta = 1$
Complete step by step Solution:
$\cos \theta \left[ {\begin{array}{*{20}{c}}
{\cos \theta }&{\sin \theta } \\
{ - \sin \theta }&{\cos \theta }
\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}
{\sin \theta }&{ - \cos \theta } \\
{\cos \theta }&{\sin \theta }
\end{array}} \right]$
We know that $a\left[ {\begin{array}{*{20}{c}}
x&y \\
z&u
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ax}&{ay} \\
{az}&{au}
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\theta }&{\sin \theta \cos \theta } \\
{ - \sin \theta \cos \theta }&{{{\cos }^2}\theta }
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{{\sin }^2}\theta }&{ - \sin \theta \cos \theta } \\
{\sin \theta \cos \theta }&{{{\sin }^2}\theta }
\end{array}} \right]$
After adding, the matrices we will get
$ = \left[ {\begin{array}{*{20}{c}}
{{{\sin }^2}\theta + {{\cos }^2}\theta }&{\sin \theta \cos \theta - \sin \theta \cos \theta } \\
{ - \sin \theta \cos \theta + \sin \theta \cos \theta }&{{{\cos }^2}\theta + {{\sin }^2}\theta }
\end{array}} \right]$
We know the identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$
$ = \left[ {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right]$
Therefore, the correct option is (D).
Note:Students should do the calculations carefully to avoid any mistakes. And use the correct identities to solve the calculations correctly. And also keep in mind that ${\cos ^2}\theta + {\sin ^2}\theta = 1$.
Formula Used: ${\cos ^2}\theta + {\sin ^2}\theta = 1$
Complete step by step Solution:
$\cos \theta \left[ {\begin{array}{*{20}{c}}
{\cos \theta }&{\sin \theta } \\
{ - \sin \theta }&{\cos \theta }
\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}
{\sin \theta }&{ - \cos \theta } \\
{\cos \theta }&{\sin \theta }
\end{array}} \right]$
We know that $a\left[ {\begin{array}{*{20}{c}}
x&y \\
z&u
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ax}&{ay} \\
{az}&{au}
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\theta }&{\sin \theta \cos \theta } \\
{ - \sin \theta \cos \theta }&{{{\cos }^2}\theta }
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{{\sin }^2}\theta }&{ - \sin \theta \cos \theta } \\
{\sin \theta \cos \theta }&{{{\sin }^2}\theta }
\end{array}} \right]$
After adding, the matrices we will get
$ = \left[ {\begin{array}{*{20}{c}}
{{{\sin }^2}\theta + {{\cos }^2}\theta }&{\sin \theta \cos \theta - \sin \theta \cos \theta } \\
{ - \sin \theta \cos \theta + \sin \theta \cos \theta }&{{{\cos }^2}\theta + {{\sin }^2}\theta }
\end{array}} \right]$
We know the identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$
$ = \left[ {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right]$
Therefore, the correct option is (D).
Note:Students should do the calculations carefully to avoid any mistakes. And use the correct identities to solve the calculations correctly. And also keep in mind that ${\cos ^2}\theta + {\sin ^2}\theta = 1$.
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