Explain Solution R. D. Sharma Class 12 Chapter deteminants Exercise 5.2 Question 2 Sub Question 15 maths Textbook Solution.

Answer:$\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\cos y \end{array}\right|=0$

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

Given: $\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\cos y \end{array}\right|$

Solution:$\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\cos y \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{R}_{3} \rightarrow-\cos y \times R_{3}-\sin y \times R_{2} \\ &=\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ (-\cos y)(-\cos x)-\sin y \sin x & (-\cos y)(\sin x)-\sin y \cos x & (-\cos y)(-\cos y)-\sin y \sin y \end{array}\right| \end{aligned}

$=\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ \cos x \cos y-\sin y \sin x & -[\sin x \cos y+\sin y \cos x] & \left(\cos ^{2} y-\sin ^{2} y\right) \end{array}\right|$

\begin{aligned} &\because \cos (x+y)=\cos x \cos y-\sin y \sin x \\ &\sin (x+y)=\sin x \cos y+\sin y \cos x \\ &\cos ^{2} y-\sin ^{2} y=\cos 2 y \end{aligned}

$=\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ \cos (x+y) & -\sin (x+y) & \cos 2 y \end{array}\right|$

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

The value of the determinant is zero

$=0 \quad\left(\because R_{1}=R_{3}\right)$

Hence $\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\cos y \end{array}\right|=0$