Precalculus Examples

$\cot^{2} (x) + \csc^{2} (x) = 2 \csc^{2} (x) - 1$

Step 1

Start on the left side.

$\cot^{2} (x) + \csc^{2} (x)$

Step 2

Apply Pythagorean identity in reverse.

$\csc^{2} (x) - 1 + \csc^{2} (x)$

Step 3

Convert to sines and cosines.

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Step 3.1

Apply the reciprocal identity to $\csc (x)$ .

${(\frac{1}{\sin (x)})}^{2} - 1 + \csc^{2} (x)$

Step 3.2

Apply the reciprocal identity to $\csc (x)$ .

${(\frac{1}{\sin (x)})}^{2} - 1 + {(\frac{1}{\sin (x)})}^{2}$

Step 3.3

Apply the product rule to $\frac{1}{\sin (x)}$ .

$\frac{1^{2}}{\sin^{2} (x)} - 1 + {(\frac{1}{\sin (x)})}^{2}$

Step 3.4

Apply the product rule to $\frac{1}{\sin (x)}$ .

$\frac{1^{2}}{\sin^{2} (x)} - 1 + \frac{1^{2}}{\sin^{2} (x)}$

Step 4

Simplify.

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Step 4.1

Combine the numerators over the common denominator.

$- 1 + \frac{1^{2} + 1^{2}}{{\sin (x)}^{2}}$

Step 4.2

Simplify each term.

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Step 4.2.1

One to any power is one.

$- 1 + \frac{1 + 1^{2}}{{\sin (x)}^{2}}$

Step 4.2.2

One to any power is one.

$- 1 + \frac{1 + 1}{{\sin (x)}^{2}}$

Step 4.3

Add $1$ and $1$ .

$- 1 + \frac{2}{\sin^{2} (x)}$

Step 5

Rewrite $- 1 + \frac{2}{\sin^{2} (x)}$ as $2 \csc^{2} (x) - 1$ .

$2 \csc^{2} (x) - 1$

Step 6

Because the two sides have been shown to be equivalent, the equation is an identity.

$\cot^{2} (x) + \csc^{2} (x) = 2 \csc^{2} (x) - 1$ is an identity