Question

$f(x) = \begin{vmatrix} \sec x & \cos x & \sec^2x + \cot x \csc x + \cos x \\ \cos^2 x & \cos^2 x & \csc^2x + \cos^4x \\ 1 & \cos^2 x & \cos^2 x \end{vmatrix}$

Answer 1

f(x)=-\sin^2x-\cos^3x.

Answer 2

We wish to evaluate

$$f(x)=\begin{vmatrix} \sec x & \cos x & \sec^2x+\cot x\,\csc x+\cos x\\[1mm] \cos^2x & \cos^2x & \csc^2x+\cos^4x\\[1mm] 1 & \cos^2x & \cos^2x \end{vmatrix}.$$

Step 1. Subtract the 2nd column from the 3rd

Replace $C_3$ by $C_3 - C_2$. Then the entries in $C_3$ become:

• Row 1: $\sec^2x+\cot x\,\csc x+\cos x - \cos x =\sec^2x+\cot x\,\csc x$.
• Row 2: $\csc^2x+\cos^4x-\cos^2x$.
  Note that $\cos^4x-\cos^2x=\cos^2x(\cos^2x-1)=-\cos^2x\sin^2x$. So this entry becomes $\csc^2x-\cos^2x\sin^2x$.
• Row 3: $\cos^2x-\cos^2x=0$.

Thus the matrix is now

$$\begin{vmatrix} \sec x & \cos x & \sec^2x+\cot x\,\csc x\\[1mm] \cos^2x & \cos^2x & \csc^2x-\cos^2x\sin^2x\\[1mm] 1 & \cos^2x & 0 \end{vmatrix}.$$

Step 2. Expand the Determinant along the 3rd Row

The 3rd row is $[1,\ \cos^2x,\ 0]$. Using cofactor expansion, we have

$$f(x)= 1\cdot M_{31} -\cos^2x\cdot M_{32}+0\cdot M_{33},$$

where

$$M_{31}=\begin{vmatrix} \cos x & \sec^2x+\cot x\,\csc x \\[1mm] \cos^2x & \csc^2x-\cos^2x\sin^2x \end{vmatrix},$$

and

$$M_{32}=\begin{vmatrix} \sec x & \sec^2x+\cot x\,\csc x \\[1mm] \cos^2x & \csc^2x-\cos^2x\sin^2x \end{vmatrix}.$$

Calculating these 2×2 determinants:

• $M_{31}=\cos x\Bigl(\csc^2x-\cos^2x\sin^2x\Bigr) - (\sec^2x+\cot x\,\csc x)\cos^2x,$
• $M_{32}=\sec x\Bigl(\csc^2x-\cos^2x\sin^2x\Bigr) - (\sec^2x+\cot x\,\csc x)\cos^2x.$

Thus

$$\begin{aligned} f(x) &=\cos x\Bigl(\csc^2x-\cos^2x\sin^2x\Bigr) - \cos^2x(\sec^2x+\cot x\,\csc x)\\[1mm] &\quad {} -\cos^2x\Bigl[\sec x\Bigl(\csc^2x-\cos^2x\sin^2x\Bigr) - (\sec^2x+\cot x\,\csc x)\cos^2x\Bigr]. \end{aligned}$$

Notice that the two parts containing $\csc^2x-\cos^2x\sin^2x$ combine as

$$\cos x - \cos^2x\sec x.$$

But since $\sec x=\frac{1}{\cos x}$, we have

$$\cos^2x\sec x=\cos^2x\cdot\frac{1}{\cos x}=\cos x.$$

Thus the terms cancel. We are left with

$$f(x) = -\cos^2x(\sec^2x+\cot x\,\csc x) + \cos^2x\cdot\cos^2x(\sec^2x+\cot x\,\csc x).$$

Factor out $\cos^2x(\sec^2x+\cot x\,\csc x)$:

$$f(x) = -\cos^2x(\sec^2x+\cot x\,\csc x)\Bigl(1-\cos^2x\Bigr).$$

Since $1-\cos^2x=\sin^2x$, we obtain

$$f(x) = -\cos^2x\,\sin^2x\Bigl(\sec^2x+\cot x\,\csc x\Bigr).$$

Step 3. Simplify the Expression in Parentheses

Recall that

$$\sec^2x=\frac{1}{\cos^2x},\qquad \cot x\,\csc x=\frac{\cos x}{\sin^2x}.$$

Thus

$$\sec^2x+\cot x\,\csc x=\frac{1}{\cos^2x}+\frac{\cos x}{\sin^2x}.$$

Plug this in:

$$f(x) = -\cos^2x\,\sin^2x\left(\frac{1}{\cos^2x}+\frac{\cos x}{\sin^2x}\right).$$

Separate the terms:

$$f(x) = -\sin^2x -\cos^2x\,\sin^2x\cdot\frac{\cos x}{\sin^2x}.$$

The second term simplifies as

$$\cos^2x\,\sin^2x\cdot\frac{\cos x}{\sin^2x}=\cos^3x.$$

Thus

$$f(x)=-\sin^2x-\cos^3x.$$