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Since $\sin$ is the reciprocal of $\csc$, $\frac{1}{\sin\left(x\right)^2\cos\left(x\right)^2}$ is equivalent to $\frac{\csc\left(x\right)^2}{\cos\left(x\right)^2}$
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$9\sec\left(x\right)^3+\csc\left(x\right)^2=\frac{\csc\left(x\right)^2}{\cos\left(x\right)^2}$
Learn how to solve problems step by step online. Solve the trigonometric equation 9sec(x)^3+csc(x)^2=1/(sin(x)^2cos(x)^2). Since \sin is the reciprocal of \csc, \frac{1}{\sin\left(x\right)^2\cos\left(x\right)^2} is equivalent to \frac{\csc\left(x\right)^2}{\cos\left(x\right)^2}. Since \cos is the reciprocal of \sec, \frac{\csc\left(x\right)^2}{\cos\left(x\right)^2} is equivalent to \csc\left(x\right)^2\sec\left(x\right)^2. Grouping all terms to the left side of the equation. Applying the trigonometric identity: \csc\left(\theta \right)^2 = 1+\cot\left(\theta \right)^2.