#### Provide solution for RD Sharma maths Class 12 Chapter 5 Determinants Exercise VSQ Question 53 maths textbook solution.

Answer : $\frac{1}{2}$

Hint: Here we use basic concept of determinant of matrix

Given: $A=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cos \theta \end{array}\right|$

Solution :

\begin{aligned} &R_{2} \rightarrow R_{2}-R_{1} \\ &A=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & \sin \theta & 0 \\ 1 & 1 & 1+\cos \theta \end{array}\right| \\ &\rightarrow R_{3} \rightarrow R_{3}-R_{1} \\ &A=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & \sin \theta & 0 \\ 0 & 0 & \cos \theta \end{array}\right| \end{aligned}

$\rightarrow$ Expanding along $C_{1}$

\begin{aligned} &C_{1}=1\left[\begin{array}{cc} \sin \theta & 0 \\ 0 & \cos \theta \end{array}\right]+0+0 \\ &=1(\sin \theta \times \cos \theta) \quad\left[\sin \theta \times \cos \theta=\frac{1}{2} \sin 2 \theta\right] \\ &=\frac{1}{2}(\sin 2 \theta) \end{aligned}

maximum value $=\frac{1}{2} \times 1=\frac{1}{2} \ldots\{\text { maximum value of } \sin \theta=1\}$