Verify the Identity tan(theta)^2cos(theta)^2=1/(csc(theta)^2)

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Trigonometry

$\tan^{2} (θ) \cos^{2} (θ) = \frac{1}{\csc^{2} (θ)}$

Start on the left side.

$\tan^{2} (θ) \cos^{2} (θ)$

Convert to sines and cosines.

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

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

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

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

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

$\sin^{2} (θ)$

Now consider the right side of the equation.

$\frac{1}{\csc^{2} (θ)}$

Convert to sines and cosines.

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

$\frac{1}{{(\frac{1}{\sin (θ)})}^{2}}$

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

$\frac{1}{\frac{1^{2}}{\sin^{2} (θ)}}$

Simplify.

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Multiply the numerator by the reciprocal of the denominator.

$1 \frac{{\sin (θ)}^{2}}{1^{2}}$

Multiply $\frac{{\sin (θ)}^{2}}{1^{2}}$ by $1$ .

$\frac{{\sin (θ)}^{2}}{1^{2}}$

One to any power is one.

$\frac{{\sin (θ)}^{2}}{1}$

Divide ${\sin (θ)}^{2}$ by $1$ .

$\sin^{2} (θ)$

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

$\tan^{2} (θ) \cos^{2} (θ) = \frac{1}{\csc^{2} (θ)}$ is an identity

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