Square of Tangent Minus Square of Sine
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 |
$\blacksquare$