Verify the Identity (cot(B) to the power of two minus cos(B) to the power of two) divide(csc(B) to the power of two minus one) equals cos(B) to the power of two

$\frac{\cot^{2} (B) - \cos^{2} (B)}{\csc^{2} (B) - 1} = \cos^{2} (B)$

Start on the left side.

$\frac{\cot^{2} (B) - \cos^{2} (B)}{\csc^{2} (B) - 1}$

Convert to sines and cosines.

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Write $\cot (B)$ in sines and cosines using the quotient identity.

$\frac{{(\frac{\cos (B)}{\sin (B)})}^{2} - \cos^{2} (B)}{\csc^{2} (B) - 1}$

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

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

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

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

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

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

Simplify.

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First, multiply the numerator and denominator of the complex fraction by ${\sin (B)}^{2}$ .

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

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

Add.

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

Use the distributive law.

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

Clarify by cancelling.

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The common factor should be canceled of ${\sin (B)}^{2}$ .

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Cancel the common factor.

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

Reformulate the expression.

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

Cancel the common factor of ${\sin (B)}^{2}$ .

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Cancel the common factor.

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

Should be rewritten the expression.

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

Clarify the numerator.

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

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Multiply by $1$ .

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

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

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

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

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

Move $- 1$ to the left of ${\sin (B)}^{2}$ .

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

Rewrite $- 1 {\sin (B)}^{2}$ as $- {\sin (B)}^{2}$ .

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

Rewrite $1 - {\sin (B)}^{2}$ in a factored form.

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

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

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

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

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

Clarify the denominator.

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One raised to any power equals one.

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

Move $- 1$ to the left of ${\sin (B)}^{2}$ .

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

Rewrite $- 1 {\sin (B)}^{2}$ as $- {\sin (B)}^{2}$ .

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

Rewrite $1 - {\sin (B)}^{2}$ in a factored form.

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

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

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

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

Cancel the common factor of $1 + \sin (B)$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $1 - \sin (B)$ .

$\cos^{2} (B)$

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

$\frac{\cot^{2} (B) - \cos^{2} (B)}{\csc^{2} (B) - 1} = \cos^{2} (B)$ is an identity

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