What is the inverse of 1/(cos^2(x)) ?

The inverse of 1/(cos^2(x)) is arccos(sqrt(1/x)),arccos(-sqrt(1/x))

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inverse of 1/(cos^2(x))

inverse

\frac{1}{cos ^{2} ( x )}

arccos (\frac{1}{x}), arccos (- \frac{1}{x})

Solution steps

Solve for

cos (x), y = \frac{1}{cos ^{2} ( x )}

cos (x) = \frac{1}{y}, cos (x) = - \frac{1}{y}

cos (a) = b \Rightarrow a = arccos (b)

arccos (\frac{1}{y}), arccos (- \frac{1}{y})

Substitute

y = x

arccos (\frac{1}{x}), arccos (- \frac{1}{x})

f (x) = \frac{3 x - 3}{- x ^{2} + 2 x - 1}

ln (x - 5)

(- 1, 1), (- 10, - 5)

f (x) = \frac{x ^{3}}{3}

f (x) = sin (\frac{x}{6})

What is the inverse of 1/(cos^2(x)) ?
The inverse of 1/(cos^2(x)) is arccos(sqrt(1/x)),arccos(-sqrt(1/x))