Square of Tangent Minus Square of Sine
From ProofWiki
Theorem
- $\tan^2 x - \sin^2 x = \tan^2 x \ \sin^2 x$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \tan^2 x - \sin^2 x\) | \(=\) | \(\displaystyle \frac {\sin^2 x} {\cos^2x} - \sin^2 x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by definition of tangent | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\sin^2 x - \sin^2 x \ \cos^2 x} {\cos^2x}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac{\sin^2 x \left({1 - \cos^2 x}\right)} {\cos^2 x}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \tan^2 x \left({1 - \cos^2 x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by definition of tangent | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \tan^2 x \ \sin^2 x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Sum of Squares of Sine and Cosine |
$\blacksquare$