Verify the Identity csc(theta)-sin(theta)=cot(theta)cos(theta)

$\csc (θ) - \sin (θ) = \cot (θ) \cos (θ)$

Start on the left side.

$\csc (θ) - \sin (θ)$

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

$\frac{1}{\sin (θ)} - \sin (θ)$

Write $- \sin (θ)$ as a fraction with denominator $1$ .

$\frac{1}{\sin (θ)} + \frac{- \sin (θ)}{1}$

Add fractions.

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

$\frac{1}{\sin (θ)} + \frac{- \sin (θ)}{1} \cdot \frac{\sin (θ)}{\sin (θ)}$

Write each expression with a common denominator of $\sin (θ)$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$\frac{1}{\sin (θ)} + \frac{- \sin (θ) \sin (θ)}{1 \sin (θ)}$

Multiply $\sin (θ)$ by $1$ .

$\frac{1}{\sin (θ)} + \frac{- \sin (θ) \sin (θ)}{\sin (θ)}$

Combine the numerators over the common denominator.

$\frac{1 - \sin (θ) \sin (θ)}{\sin (θ)}$

Multiply $- \sin (θ) \sin (θ)$ .

$\frac{1 - \sin^{2} (θ)}{\sin (θ)}$

Apply pythagorean identity.

$\frac{\cos^{2} (θ)}{\sin (θ)}$

Now consider the right side of the equation.

$\cot (θ) \cos (θ)$

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

$\frac{\cos (θ)}{\sin (θ)} \cos (θ)$

Multiply $\frac{\cos (θ)}{\sin (θ)} \cos (θ)$ .

$\frac{\cos^{2} (θ)}{\sin (θ)}$

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

$\csc (θ) - \sin (θ) = \cot (θ) \cos (θ)$ is an identity

Verify the Identity csc(theta)-sin(theta)=cot(theta)cos(theta)

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