Square of Tangent Minus Square of Sine

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Theorem

$\tan^2 x - \sin^2 x = \tan^2 x \ \sin^2 x$


Proof

\(\ds \tan^2 x - \sin^2 x\) \(=\) \(\ds \frac {\sin^2 x} {\cos^2x} - \sin^2 x\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds \frac {\sin^2 x - \sin^2 x \ \cos^2 x} {\cos^2x}\)
\(\ds \) \(=\) \(\ds \frac{\sin^2 x \left({1 - \cos^2 x}\right)} {\cos^2 x}\)
\(\ds \) \(=\) \(\ds \tan^2 x \left({1 - \cos^2 x}\right)\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds \tan^2 x \ \sin^2 x\) Sum of Squares of Sine and Cosine

$\blacksquare$