Solve 1/(cos^2x*sin^2x)=sec^2x csc^2x

Evaluate

\frac{1}{cos ^{2} ( x ) sin ^{2} ( x )} = sec^{2} (x) csc^{2} (x)

Find the domain

More Steps

\frac{1}{cos ^{2} ( x ) sin ^{2} ( x )} = sec^{2} (x) csc^{2} (x), x \neq = \frac{kπ}{2}, k \in Z

Multiply the terms

\frac{1}{( cos ( x ) sin ( x ) ) ^{2}} = sec^{2} (x) csc^{2} (x)

Multiply the terms

\frac{1}{( cos ( x ) sin ( x ) ) ^{2}} = (sec (x) csc (x))^{2}

Rewrite the expression

\frac{1}{( cos ( x ) sin ( x ) ) ^{2}} = (\frac{1}{cos ( x )} \times \frac{1}{sin ( x )})^{2}

Expand the expression

\frac{1}{cos ^{2} ( x ) sin ^{2} ( x )} = cos^{- 2} (x) sin^{- 2} (x)

Calculate

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

Cross multiply

(cos (x) sin (x))^{2} = cos^{2} (x) sin^{2} (x)

Move the expression to the left side

(cos (x) sin (x))^{2} - cos^{2} (x) sin^{2} (x) = 0

Calculate

More Steps

0 = 0

The statement is true for any value of x

x \in R

Check if the solution is in the defined range

x \in R, x \neq = \frac{kπ}{2}, k \in Z

Solution

x \neq = \frac{kπ}{2}, k \in Z

Alternative Form

x \neq = 9 0^{\circ} k, k \in Z

Solve the equation