Question

Let
$f\left(x\right)=\left|\begin{array}{ccc}{\sec }^{2}x& 1& 1\\ {\cos }^{2}x& {\cos }^{2}x& cose{c}^{2}x\\ 1& {\cos }^{2}x& {\cot }^{2}x\end{array}\right|$
then

Answer 1

$f\left(x\right)=\left|\begin{array}{ccc}{\sec }^{2}x& 1& 1\\ {\cos }^{2}x& {\cos }^{2}x& cose{c}^{2}x\\ 1& {\cos }^{2}x& {\cot }^{2}x\end{array}\right|$
Applying $C_2\to C_2-{\cos }^{2}x{C}_1$, we get
$f\left(x\right)=\left|\begin{array}{ccc}{\sec }^{2}x& 0& 1\\ {\cos }^{2}x& {\cos }^{2}x-{\cos }^{4}x& cose{c}^{2}x\\ 1& 0& {\cot }^{2}x\end{array}\right|=({\cos }^{2}x-{\cos }^{4}x)({\sec }^{2}x{\cot }^{2}x-1)={\cos }^{2}x{\sin }^{2}x({\csc }^{2}x-1)={\cos }^{2}x{\sin }^{2}x{\cot }^{2}x={\cos }^{4}x$
A) Period of ${\cos }^{4}x$ is $\pi$
B) As $0\le {\cos }^{4}x\le 1$ maximum value of $f\left(x\right)$ is $1$
C) ${\int }_0^{\pi /4}f\left(x\right)dx-\frac{1}{4}={\int }_0^{\pi /4}\frac{1}{8}(3+4\cos 2x+\cos 4x)dx-\frac{1}{4}$
$=\frac{1}{4}{[3x+2\sin 2x+\frac{1}{4}\sin 4x]}_0^{\pi /4}-\frac{1}{4}=\frac{1}{8}(\frac{3\pi }{4}+2)-\frac{1}{4}=\frac{3\pi }{32}$
D) As $0\le {\cos }^{4}x\le 1$ minimum values of $f\left(x\right)$ is $0$