#### need solution for RD Sharma maths class 12 chapter 5 Determinants exercise Fill in the blanks question 35

Hint: Here, we use basic concept of determinant of matrix

Given: $\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ -1 & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right]$

Solution:

$\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ -1 & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right]$

\begin{aligned} &{\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2}\left(90^{\circ}-23^{\circ}\right) & -1 \\ -\sin ^{2}\left(90^{\circ}-23^{\circ}\right) & -\sin ^{2} 23^{\circ} & 1 \\ -1 & \sin ^{2} 23^{\circ} & \cos ^{2}\left(90^{\circ}-23^{\circ}\right) \end{array}\right]} \\ &{\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \cos ^{2} 23 & -1 \\ -\cos ^{2} 23 & -\sin ^{2} 23^{\circ} & 1 \\ -1 & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right]} \end{aligned}

Let's apply $C_{1} \rightarrow C_{1}+C_{2}$

\begin{aligned} &{\left[\begin{array}{ccc} \sin ^{2} 23^{\circ}+\cos ^{2} 23^{\circ} & \cos ^{2} 23 & -1 \\ -\cos ^{2} 23-\sin ^{2} 23^{\circ} & -\sin ^{2} 23^{\circ} & 1 \\ -1+\sin ^{2} 23^{\circ} & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right]} \\ &{\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & -\sin ^{2} 23^{\circ} & 1 \\ -\cos ^{2} 23^{\circ} & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right]} \\ &(-1)\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & -\sin ^{2} 23^{\circ} & 1 \\ -\cos ^{2} 23^{\circ} & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right] \end{aligned}

So, here two columns become same. So, determinant value becomes zero.