Trigonometry Examples

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

Step 1

Start on the left side.

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

Step 2

Simplify each term.

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Step 2.1

Rewrite $1$ as $1^{2}$ .

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

Step 2.2

Rewrite $\frac{1^{2}}{\tan^{2} (x)}$ as ${(\frac{1}{\tan (x)})}^{2}$ .

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

Step 2.3

Rewrite $\tan (x)$ in terms of sines and cosines.

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

Step 2.4

Multiply by the reciprocal of the fraction to divide by $\frac{\sin (x)}{\cos (x)}$ .

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

Step 2.5

Write $1$ as a fraction with denominator $1$ .

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

Step 2.6

Simplify.

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Step 2.6.1

Rewrite the expression.

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

Step 2.6.2

Multiply $\frac{\cos (x)}{\sin (x)}$ by $1$ .

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

Step 2.7

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

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

Step 3

Apply Pythagorean identity in reverse.

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Rewrite $1$ as $1^{2}$ .

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

Step 4.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = \sin (x)$ .

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

Step 4.2

Write $1$ as a fraction with a common denominator.

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

Step 4.3

Combine the numerators over the common denominator.

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

Step 4.4

Simplify the numerator.

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

Step 5

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

$1 + \frac{1}{\tan^{2} (x)} = \frac{1}{\sin^{2} (x)}$ is an identity