The Inverse of a Matrix

Show that the matrix is invertible(nonsingular) and find its inverse:

A = sec θ tan θ
tan θ sec θ

*just imagine that is a 2x2 matrix

*need your help I'm stuck with this problem, I'm weak in solving problems involving trigonometric functions :(

It is always a good idea to ask if the matrix is singular, before you try to invert it. For a 2x2 matrix, that is best done by taking the determinant.

det(A) = sec^2(θ) - tan^2(θ) = 1

This comes from a basic identity between the sec and tan functions, that

sec^2(θ) = 1 + tan^2(θ)

(It is very worthwhile memorizing these basic identities, as you will use them over and over. While you can re-derive them in theory, that takes time and wastes mental energy.) So the determinant of A is non-singular for ALL values of θ.

Next, computing the inverse matrix is distressingly simple for a 2x2 matrix once you have the determinant. You simply compute the adjugate matrix (the transpose of the matrix of cofactors), and then divide by the determinant. But the determinant of A is already known to be 1, and the matrix is symmetric and only 2x2, so the matrix of cofactors is simple.

inv(A) = adj(A)/det(A) = adj(A)/1 =
[ sec(θ) -tan(θ) ]
[-tan(θ) sec(θ) ]

determinant = sec²θ - tan²θ
but sec²θ - tan²θ = 1/cos²θ - sin²θ/cos²θ = cos²θ/cos²θ = 1

so the determinant is nonzero and is thus invertible

then A^-1 is
[secθ -tanθ]
[-tanθ secθ]