#### Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 28 maths

Correct option (a)

Hint:

First solve the given determinant then find the maximum value of sin function, which is 1.

Given:

Given that,

$\Delta =\begin{vmatrix} 1 &1 &1 \\ 1 &1+sin\, \theta &1 \\ 1+cos\, \theta &1 &1 \end{vmatrix}$

We have to find the maximum value of $\Delta$.

Solution:

$Here\: \: \Delta =\begin{vmatrix} 1 &1 &1 \\ 1 &1+sin\, \theta &1 \\ 1+cos\, \theta &1 &1 \end{vmatrix}$

Applying R1 → R1 - R2

$\Rightarrow \Delta =\begin{vmatrix} 1 &1 &1 \\ 1 &1+sin\, \theta &1 \\ 1+cos\, \theta &1 &1 \end{vmatrix}$

Expanding along R1

\begin{aligned} &\Rightarrow \quad \Delta=\sin \theta[1-1-\cos \theta] \\ &\Rightarrow \quad \Delta=-\sin \theta \cos \theta \\ &\Rightarrow \quad \Delta=\frac{-2 \sin \theta \cos \theta}{2} \\ &\Rightarrow \quad \Delta=\frac{-\sin 2 \theta}{2} \end{aligned}            $[\because \sin \sin 2 \theta=2 \sin \sin \theta \cos \cos \theta]$

We know that the maximum value of sin function is 1.

Hence, maximum value of

$\Delta =\frac{1}{2}$