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

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \tan^2 x - \sin^2 x\) \(=\) \(\displaystyle \frac {\sin^2 x} {\cos^2x} - \sin^2 x\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          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          

$\blacksquare$

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