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${\mathrm{tan}}^{2}\left(\alpha \right)-{\mathrm{sin}}^{2}\left(\alpha \right)={\mathrm{tan}}^{2}\left(\alpha \right){\mathrm{sin}}^{2}\left(\alpha \right)$ Start on the left side.

${\mathrm{tan}}^{2}\left(\alpha \right)-{\mathrm{sin}}^{2}\left(\alpha \right)$

Convert to sines and cosines.

Write $\mathrm{tan}\left(\alpha \right)$ in sines and cosines using the quotient identity.

${\left(\frac{\mathrm{sin}\left(\alpha \right)}{\mathrm{cos}\left(\alpha \right)}\right)}^{2}-{\mathrm{sin}}^{2}\left(\alpha \right)$

Apply the product rule to $\frac{\mathrm{sin}\left(\alpha \right)}{\mathrm{cos}\left(\alpha \right)}$.

$\frac{{\mathrm{sin}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}-{\mathrm{sin}}^{2}\left(\alpha \right)$

Write $-{\mathrm{sin}}^{2}\left(\alpha \right)$ as a fraction with denominator $1$.

$\frac{{\mathrm{sin}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}+\frac{-{\mathrm{sin}}^{2}\left(\alpha \right)}{1}$

To write $\frac{-{\mathrm{sin}}^{2}\left(\alpha \right)}{1}$ as a fraction with a common denominator, multiply by $\frac{{\mathrm{cos}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$.

$\frac{{\mathrm{sin}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}+\frac{-{\mathrm{sin}}^{2}\left(\alpha \right)}{1}\cdot \frac{{\mathrm{cos}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$

Multiply $\frac{-{\mathrm{sin}}^{2}\left(\alpha \right)}{1}$ by $\frac{{\mathrm{cos}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$.

$\frac{{\mathrm{sin}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}+\frac{-{\mathrm{sin}}^{2}\left(\alpha \right){\mathrm{cos}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$

Combine the numerators over the common denominator.

$\frac{{\mathrm{sin}}^{2}\left(\alpha \right)-{\mathrm{sin}}^{2}\left(\alpha \right){\mathrm{cos}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$

Apply Pythagorean identity.

Multiply by $1$.

$\frac{{\mathrm{sin}}^{2}\left(\alpha \right)\cdot 1-{\mathrm{sin}}^{2}\left(\alpha \right){\mathrm{cos}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$

Factor ${\mathrm{sin}}^{2}\left(\alpha \right)$ out of $-{\mathrm{sin}}^{2}\left(\alpha \right){\mathrm{cos}}^{2}\left(\alpha \right)$.

$\frac{{\mathrm{sin}}^{2}\left(\alpha \right)\cdot 1+{\mathrm{sin}}^{2}\left(\alpha \right)\left(-{\mathrm{cos}}^{2}\left(\alpha \right)\right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$

Factor ${\mathrm{sin}}^{2}\left(\alpha \right)$ out of ${\mathrm{sin}}^{2}\left(\alpha \right)\cdot 1+{\mathrm{sin}}^{2}\left(\alpha \right)\left(-{\mathrm{cos}}^{2}\left(\alpha \right)\right)$.

$\frac{{\mathrm{sin}}^{2}\left(\alpha \right)\cdot \left(1-{\mathrm{cos}}^{2}\left(\alpha \right)\right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$

Apply pythagorean identity.

$\frac{{\mathrm{sin}}^{2}\left(\alpha \right)\cdot {\mathrm{sin}}^{2}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$

Multiply${\mathrm{sin}\left(\alpha \right)}^{2}$ by ${\mathrm{sin}\left(\alpha \right)}^{2}$ by adding the exponents.

$\frac{{\mathrm{sin}}^{4}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$

Rewrite $\frac{{\mathrm{sin}}^{4}\left(\alpha \right)}{{\mathrm{cos}}^{2}\left(\alpha \right)}$ as ${\mathrm{tan}}^{2}\left(\alpha \right){\mathrm{sin}}^{2}\left(\alpha \right)$.

${\mathrm{tan}}^{2}\left(\alpha \right){\mathrm{sin}}^{2}\left(\alpha \right)$

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

${\mathrm{tan}}^{2}\left(\alpha \right)-{\mathrm{sin}}^{2}\left(\alpha \right)={\mathrm{tan}}^{2}\left(\alpha \right){\mathrm{sin}}^{2}\left(\alpha \right)$ is an identity