We need to find the maximum value of
\(\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+sin\,\theta & 1 \\ 1 & 2 & 1+cos\,\theta \end{vmatrix}\)
Let us find the determinant,
\(\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+sin\,\theta & 1 \\ 1 & 2 & 1+cos\,\theta \end{vmatrix}\)
Determinant of 3 × 3 matrices is found as,
So,
Multiply and divide by 2 on right hand side,
[∵, By trigonometric identity, sin 2θ = 2 sin θ cos θ]
We need to find the maximum value of \(\frac{sin\,2\theta}{2}.\)
We know the range of sine function.
-1 ≤ sin A ≤ 1
Or,
-1 ≤ sin 2θ ≤ 1
∴, maximum value of sin 2θ is 1.
\(\Rightarrow\) maximum value of \(\frac{sin\,2\theta}{2}=1/2\)
Thus, maximum value of
\(\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+sin\,\theta & 1 \\ 1 & 2 & 1+cos\,\theta \end{vmatrix}=\frac{1}{2}\)
