Question Statement:-
If $u_n=\displaystyle\int^{\frac{\pi}{2}}_{0}{\dfrac{1-\cos{2nx}}{1-\cos{2x}}dx}$, then find the value of the determinant $$\Delta=\begin{vmatrix} \dfrac{\pi}{2} & u_2 & u_3 \\ u_4 & u_5 & u_6 \\ u_7 & u_8 & u_9 \end{vmatrix}$$
My Solution:-
$u_1=\displaystyle\int^{\frac{\pi}{2}}_{0}{\dfrac{1-\cos{2x}}{1-\cos{2x}}dx}=\dfrac{\pi}{2}$
$\therefore$ The given determinant becomes $$\Delta=\begin{vmatrix} u_1 & u_2 & u_3 \\ u_4 & u_5 & u_6 \\ u_7 & u_8 & u_9 \end{vmatrix}$$
Now, at this point I considered it wise to use to use some row and column operations to solve the determinant rather than expanding the determinant. So for coming up with a good enough row operation that gives a constant, what I considered doing was to use the property $\cos^2{m}-\cos^{2}n=\sin{(n+m)}\cdot\sin{(n-m)}$ to my advantage.
As, $u_n=\displaystyle\int^{\frac{\pi}{2}}_{0}{\dfrac{1-\cos{2nx}}{1-\cos{2x}}dx}=\displaystyle\int^{\frac{\pi}{2}}_{0}{\dfrac{\sin^2{nx}}{\sin^2{x}}dx}$
$\therefore u_{n+1}-u_n+u_{n+1}-u_{n+2} =\displaystyle\int^{\frac{\pi}{2}}_{0}{\dfrac{\left(\sin^2{(n+1)}x-\sin^2{(n)}x\right)+\left(\sin^2{(n+1)}x-\sin^2{(n+2)}x\right)}{\sin^2{x}}dx} =\displaystyle\int^{\frac{\pi}{2}}_{0}{\dfrac{\left(\sin(2n+1)x\sin{x}\right)+\left(\sin(2n+3)x\sin(-x)\right)}{\sin^2{x}}dx} =\displaystyle\int^{\frac{\pi}{2}}_{0}{\dfrac{\sin{x}\left(\sin(2n+1)x-\sin(2n+3)x\right)}{\sin^2{x}}dx} =\displaystyle\int^{\frac{\pi}{2}}_{0}{\dfrac{2\sin^2{x}\cos2(n+1)x}{\sin^2{x}}dx} =\displaystyle\int^{\frac{\pi}{2}}_{0}{\left(2\cos2(n+1)x\right)dx}=\left(\dfrac{\sin2(n+1)x}{(n+1)}\right)_{0}^{\frac{\pi}{2}}=0$
So, we get $$\boxed{u_n+u_{n+2}-2u_{n+1}=0}$$
$\therefore$ we apply the column operation $C_1\rightarrow C_1+C_3-2C_2$ to $\Delta$ to get
$$\Delta=\begin{vmatrix} 0 & u_2 & u_3 \\ 0 & u_5 & u_6 \\ 0 & u_8 & u_9 \end{vmatrix}=0$$
While writing my solution I figured out the mistake that I was making but since I had already written all the above I just couldn't make my heart bleed by not posting all my work, so instead of asking "how to proceed further" or "where I am going wrong" I will be asking - Is there any better way to solve this question at highschool level?
And, by the way the mistake that I had been making was to write $1-\cos{2nx}=\cos^2{nx}$ and $1-\cos{2x}=\cos^2{x}$, typical isn't it until you realise that I spent a whole day trying to solve this question and not noticing this stupid mistake.
The determinant is zero by the cosine addition formulas:
$$u_{n-1}-2u_n+u_{n+1} = \int_{0}^{\pi/2}\frac{2\cos(2nx)-\cos((2n-2)x)-\cos((2n+2)x)}{1-\cos(2x)}\,dx $$ leads to: $$ u_{n-1}-2u_n + u_{n+1} = \int_{0}^{\pi/2}2\cos(2n x)\,dx = 0.$$ It follows that the second column of the given matrix is a linear combination of the first column and the third one.