Determine if True tan(x)^2-sin(x)^2=tan(x)^2sin(x)^2

$\tan^{2} (x) - \sin^{2} (x) = \tan^{2} (x) \sin^{2} (x)$

Step 1

Start on the left side.

$\tan^{2} (x) - \sin^{2} (x)$

Step 2

Convert to sines and cosines.

Write $\tan (x)$ in sines and cosines using the quotient identity.

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

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

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

Step 3

Write $- \sin^{2} (x)$ as a fraction with denominator $1$ .

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

Step 4

Add fractions.

To write $\frac{- \sin^{2} (x)}{1}$ as a fraction with a common denominator, multiply by $\frac{\cos^{2} (x)}{\cos^{2} (x)}$ .

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

Multiply $\frac{- \sin^{2} (x)}{1}$ by $\frac{\cos^{2} (x)}{\cos^{2} (x)}$ .

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

Combine the numerators over the common denominator.

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

Step 5

Apply Pythagorean identity.

Multiply by $1$ .

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

Factor $\sin^{2} (x)$ out of $- \sin^{2} (x) \cos^{2} (x)$ .

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

Factor $\sin^{2} (x)$ out of $\sin^{2} (x) \cdot 1 + \sin^{2} (x) (- \cos^{2} (x))$ .

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

Apply pythagorean identity.

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

Step 6

Multiply ${\sin (x)}^{2}$ by ${\sin (x)}^{2}$ by adding the exponents.

$\frac{\sin^{4} (x)}{\cos^{2} (x)}$

Step 7

Rewrite $\frac{\sin^{4} (x)}{\cos^{2} (x)}$ as $\tan^{2} (x) \sin^{2} (x)$ .

$\tan^{2} (x) \sin^{2} (x)$

Step 8

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

$\tan^{2} (x) - \sin^{2} (x) = \tan^{2} (x) \sin^{2} (x)$ is an identity