explain solution RD Sharma class 12 chapter 5 Determinants exercise Fill in the blanks question 32 maths

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

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

Given: $\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cot \theta \end{array}\right]$

Solution:

$\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cot \theta \end{array}\right]$

\begin{aligned} &C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow C_{3}-1 \\ &\Delta=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & \sin \theta & 0 \\ 1 & 0 & \cot \theta \end{array}\right] \\ &1\left[\begin{array}{cc} \sin \theta & 0 \\ 0 & \cot \theta \end{array}\right]-0\left[\begin{array}{cc} 1 & 0 \\ 1 & \cot \theta \end{array}\right]+0\left[\begin{array}{cc} 1 & \sin \theta \\ 1 & 0 \end{array}\right] \end{aligned}

\begin{aligned} &1(\sin \theta \times \cot \theta)-0(0)+0(0) \\ &\sin \theta \times \cot \theta-0+0 \\ &\sin \theta \times \frac{\cos \theta}{\sin \theta}=\cos \theta \\ &\Delta=\cos \theta \end{aligned}

Let's $\Delta =0$

$\cos \; \theta =0=\frac{\pi }{2} \left [ So,maximum\; value\; is\; \frac{1}{2}\; of \; \cos \theta \right ]$