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 |
$\blacksquare$