# $\text{Prove that the determinant } \begin{vmatrix} x&sin\theta&cos\theta \\ -sin\theta&-x&1\\ cos\theta&1&x \end{vmatrix} \text{ is independent of } \theta$

Toolbox:
• A determinant can be expanded along the rows or any of the column with their corresponding cofactors,to find its value.
We have to prove that the given determinant is independent of $\theta$.That is,it has only algebraic terms.
Given $\bigtriangleup=\begin{vmatrix} x&sin\theta&cos\theta \\ -sin\theta&-x&1\\ cos\theta&1&x \end{vmatrix}$
Now let us expand along $R_1$
$x(-x\times x-1\times 1)-sin\theta(\sin\theta\times x-1\times cos\theta)+cos\theta(-sin\theta\times 1-cos\theta\times -x)$
On expanding we get,
$=x(-x^2-1)-sin\theta(-xsin\theta-cos\theta)+\cos\theta(-sin\theta+xcos\theta)$
On simplifying we get,
=$-x^3-x+xsin^2\theta+sin\theta cos\theta-sin\theta cos\theta-x cos^2\theta$
=$-x^3-x+x(sin^2\theta+cos^2\theta)$
But we know $(sin^2\theta+cos^2\theta)=1$
Therefore $\bigtriangleup=-x^3-x+x$.
Therefore $\bigtriangleup=-x^3$.
Which shows that $\bigtriangleup$ is only algebraic and is independent of $\theta$.