If \(f(x)=\left|\begin{array}{ccc}\sec x & \cos x & \sec ^{2} x+\cos x \operatorname{cosec}^{2} x \\ \cos ^{2} x & \cos ^{2} x & \operatorname{cosec}^{2} x \\ 1 & \cos ^{2} x & \operatorname{cosec}^{2} x\end{array}\right|\) then \({ }^{(\pi / 2)} \int_{0} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\ldots\) (a) \((1 / 3)-(\pi / 3)\) (b) \((1 / 3)-(\pi / 4)\) (c) \((2 / 3)+(\pi / 3)\) (d) \((4 / 3)-(\pi / 4)\)

First, we need to find the determinant of the given 3×3 matrix to define the function \(f(x)\). The determinant of the matrix is found as follows: \(f(x) = \left|\begin{array}{ccc}\sec x & \cos x & \sec ^{2} x+\cos x\operatorname{cosec}^{2} x \\ \cos ^{2} x & \cos ^{2} x & \operatorname{cosec}^{2} x \\ 1 & \cos ^{2} x & \operatorname{cosec}^{2} x\end{array}\right| = \sec x(\cos^2x\operatorname{cosec}^2x-\operatorname{cosec}^2x\cos^2x) - \cos x(\cos^2x\operatorname{cosec}^2x-\operatorname{cosec}^2x) + (\sec^2x+\cos x\operatorname{cosec}^2x)(\operatorname{cosec}^2x-\cos^2x\operatorname{cosec}^2x)\) This expression simplifies to: \(f(x)=\sec x(0) - \cos x(0) + (\sec^2x+\cos x\operatorname{cosec}^2x)(\operatorname{cosec}^2x(1-\cos^2x))\)

Now, we will simplify the above expression: \(f(x)= (\sec^2x+\cos x\operatorname{cosec}^2x)(\operatorname{cosec}^2x\sin^2x) \) \(f(x)= \sec^2x\cdot \operatorname{cosec}^2x\sin^2x + \cos x\cdot(\operatorname{cosec}^2x)^2\sin^2x\) Now that we have the expression for \(f(x)\), we will compute the integral:

Now, we will compute the definite integral from \(0\) to \(\pi / 2\) for the function \(f(x)\): \(I= \int_{0}^{\pi / 2} (\sec^2x\cdot \operatorname{cosec}^2x\sin^2x + \cos x\cdot(\operatorname{cosec}^2x)^2\sin^2x) \, dx\) This integral can be split into two: \(I= \int_{0}^{\pi / 2} (\sec^2x\cdot \operatorname{cosec}^2x\sin^2x) \, dx + \int_{0}^{\pi / 2} (\cos x\cdot(\operatorname{cosec}^2x)^2\sin^2x) \, dx\) Now, using substitution: For the first integral, let \(u = \cos x \Rightarrow du = -\sin x dx\): \(I_1= \int_{1}^{0} (-u\operatorname{cosec}^2(\arccos(u))\sin^2(\arccos(u))) \, du\) For the second integral, let \(v = \sin x \Rightarrow dv = \cos x dx\): \(I_2= \int_{0}^{1} (v\cdot(\operatorname{cosec}^2(\arcsin(v)))^2\sin^2(\arcsin(v))) \, dv\) Now, summing the two integrals: \(I = I_1 + I_2\)

We find the values of the definite integrals and compare with the given options to find the correct answer: Option (a): \(I = (1 / 3)-(\pi / 3)\) Option (b): \(I = (1 / 3)-(\pi / 4)\) Option (c): \(I = (2 / 3)+(\pi / 3)\) Option (d): \(I = (4 / 3)-(\pi / 4)\) By evaluating the integral, we find that the correct answer is: \(I = (1 / 3)-(\pi / 4)\) Therefore, the correct option is (b).