Question:
I know how to solve this problem but I need a faster approach to this problem.
By using properties of determinants prove that the determinant $$\begin{vmatrix}a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a\end{vmatrix}$$ is independent of $x$.
My Approach:
\begin{align*} \text{LHS} &= \begin{vmatrix}a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a\end{vmatrix} \\&= -a^3 \begin{vmatrix}1 & \dfrac{\sin x}{a} & \dfrac{\cos x}{a} \\ \dfrac{\sin x}{a} & a & \dfrac{-1}{a} \\ \dfrac{\cos x}{a} & \dfrac{1}{a} & 1\end{vmatrix}\begin{array}{l}\text{Taking a common $a$ from $R_1$}\\\text{Taking a common $-a$ from $R_2$}\\\text{Taking a common $a$ from $R_3$}\end{array}\end{align*}
If you use the "criss-cross" method for a $3\times3$ determinant $$\left|\matrix{a&b&c\cr d&e&f\cr g&h&i\cr}\right|=aei+bfg+cdh-ceg-afh-bdi$$ you can just write down the answer: $$\eqalign{\det(\hbox{your matrix}) &=-a^3+\cos x\sin x-\sin x\cos x+a\cos^2x+a\sin^2x-a\cr &=-a^3\ .\cr}$$ I really doubt you'll get anything faster than that.
If you haven't seen the "criss-cross" method before, it is easy to remember: copy out the first two columns again, add up these three products $$\def\c#1{\color{red}{#1}} \left|\matrix{\c a&b&c\cr d&\c e&f\cr g&h&\c i\cr}\right|\matrix{a&b\cr d&e\cr g&h\cr}\qquad \left|\matrix{a&\c b&c\cr d&e&\c f\cr g&h&i\cr}\right|\matrix{a&b\cr d&e\cr \c g&h\cr}\qquad \left|\matrix{a&b&\c c\cr d&e&f\cr g&h&i\cr}\right|\matrix{a&b\cr \c d&e\cr g&\c h\cr}$$ and subtract these three products $$\def\c#1{\color{blue}{#1}} \left|\matrix{a&b&\c c\cr d&\c e&f\cr \c g&h&i\cr}\right|\matrix{a&b\cr d&e\cr g&h\cr}\qquad \left|\matrix{a&b&c\cr d&e&\c f\cr g&\c h&i\cr}\right|\matrix{\c a&b\cr d&e\cr g&h\cr}\qquad \left|\matrix{a&b&c\cr d&e&f\cr g&h&\c i\cr}\right|\matrix{a&\c b\cr \c d&e\cr g&h\cr}$$ But NOTE that it only works for the $3\times3$ case.