#### Please solve RD Sharma class 12 chapter Determinants exercise multiple choise question 21 maths textbook solution

Correct option (d)

Hint:

First solve the given determinant then find |A| for $\Theta$ =0 and for

$\theta =\frac{\pi }{2},\frac{3\pi }{2},.......$

Given:

Given that,

$Let\; \; A=\begin{vmatrix} 1 &sin\theta &1 \\ -sin\theta &1 &sin\theta \\ -1 &-sin\theta &1 \end{vmatrix}$

$W\! here\; \; 0\leq \theta \leq 2\pi$

We have to find the |A|

Solution:

$Here\; \; A=\begin{vmatrix} 1 &sin\theta &1 \\ -sin\theta &1 &sin\theta \\ -1 &-sin\theta &1 \end{vmatrix}$

Expanding along R1, we get

$\left | A \right |=1+sin^{2}\theta -sin\theta (-sin\theta +sin\theta )+sin^{2}\theta +1$

$\Rightarrow \left | A \right |=2+2sin^{2}$

$\Rightarrow \left | A \right |=2(1+\theta )$

$Given\; that\; \; 0\leq \theta \leq 2\pi$

For $\theta$=0

$\Rightarrow \left | A \right |=2$

$F\! or\; \theta =\frac{\pi }{2},\frac{3\pi }{2},.......$

$\Rightarrow \left | A \right |=2(1+1)$

$\Rightarrow \left | A \right |=4$

$Hence \left | A \right |\varepsilon [2,4]$