Verify the Identity sec(x)tan(x)(1-sin(x)^2)=sin(x)

$\sec (x) \tan (x) (1 - \sin^{2} (x)) = \sin (x)$

Start on the left side.

$\sec (x) \tan (x) (1 - \sin^{2} (x))$

Apply pythagorean identity.

$\sec (x) \tan (x) \cos^{2} (x)$

Convert to sines and cosines.

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

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

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

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

Simplify.

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

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

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

Raise $\cos (x)$ to the power of $1$ .

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

Raise $\cos (x)$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Cancel the common factor of ${\cos (x)}^{2}$ .

$\sin (x)$

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

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

Verify the Identity sec(x)tan(x)(1-sin(x)^2)=sin(x)

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