#### Explain Solution R. D. Sharma Class 12 Chapter 5.2 deteminants Exercise MCQs Question 2 Sub Question 14 maths Textbook Solution.

Answer:$\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ \cos 180^{\circ} & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right|=0$

Hint: We will try to do any two column or row equal

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

Solution:$\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ \cos 180^{\circ} & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right|$

\begin{aligned} &=\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \cos ^{2}\left(90^{\circ}-67^{\circ}\right) & -1 \\ -\sin ^{2} 67^{\circ} & -\cos ^{2}\left(90^{\circ}-23^{\circ}\right) & (-1)^{2} \\ \cos 180^{\circ} & \cos ^{2}\left(90^{\circ}-23^{\circ}\right) & \sin ^{2} 67^{\circ} \end{array}\right| \\ &\because \cos (90-\theta)=\sin \theta \\ &\cos 120^{\circ}=-1 \\ &=\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} & -1 \\ -\sin ^{2} 67^{\circ} & -\cos ^{2} 67^{0} & 1 \\ -1 & \cos ^{2} 67^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right| \end{aligned}

\begin{aligned} &C_{1} \rightarrow C_{1}+C_{2} \\ &=\left|\begin{array}{ccc} \sin ^{2} 23^{\circ}+\cos ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} & -1 \\ -\sin ^{2} 67^{\circ}+\left(-\cos ^{2} 67^{\circ}\right) & -\cos ^{2} 67^{\circ} & 1 \\ -1+\cos ^{2} 67^{\circ} & \cos ^{2} 67^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right| \\ &=\left|\begin{array}{ccc} 1 & \cos ^{2} 23^{\circ} & -1 \\ -\left(\sin ^{2} 67^{\circ}+\cos ^{2} 67^{\circ}\right) & -\cos ^{2} 67^{\circ} & 1 \\ -1+\cos ^{2} 67^{0} & \cos ^{2} 67^{0} & \sin ^{2} 67^{\circ} \end{array}\right| \\ &\because \sin ^{2} \theta+\cos ^{2} \theta=1 \end{aligned}

\begin{aligned} &=\left|\begin{array}{ccc} 1 & \cos ^{2} 23^{0} & -1 \\ -1 & -\cos ^{2} 67^{\circ} & 1 \\ -\sin ^{2} 67^{\circ} & \cos ^{2} 67^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right|\\ &\text { Taking common }(-1) \text { from } \mathrm{C}_{1}\\ &=(-1)\left|\begin{array}{ccc} -1 & \cos ^{2} 23^{0} & -1 \\ 1 & -\cos ^{2} 67^{\circ} & 1 \\ \sin ^{2} 67^{\circ} & \cos ^{2} 67^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right| \end{aligned}

If any two rows or columns of a determinant are identical.

The value of the determinant is zero

\begin{aligned} &=(-1) \times 0 \\ &=0 \end{aligned}

Hence $\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ \cos 180^{\circ} & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right|=0$