Verify the Identity (cot(B)^2-cos(B)^2)/(csc(B)^2-1)=cos(B)^2

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Trigonometry

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

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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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 $\frac{{\sin (B)}^{2}}{{\sin (B)}^{2}}$ .

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

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

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Cancel the common factor.

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

Rewrite 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.

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

Rewrite the expression.

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

Simplify 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 $1^{2}$ .

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

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 $b = \sin (B)$ .

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

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

Simplify the denominator.

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One to any power is 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 $1^{2}$ .

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

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 $b = \sin (B)$ .

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

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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