Prove-that-tan-2A-2tan-A-1-tan-2-A-

Question Number 130221 by bramlexs22 last updated on 23/Jan/21

Prove that \tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A}

Answered by EDWIN88 last updated on 23/Jan/21

by {De}^{'} Moivre theorem

\cos 2 A + i \sin 2 A = {(\cos A + i \sin A)}^{2}

= (\cos^{2} A - \sin^{2} A) + i (2 \sin Acos A)

\frac{\cos 2 A + i \sin 2 A}{\cos 2 A} = \frac{\cos^{2} A - \sin^{2} A}{\cos 2 A} + i \frac{2 \sin Acos A}{\cos 2 A}

1 + i \tan 2 A = 1 + i \frac{2 \sin Acos A}{\cos^{2} A - \sin^{2} A}

\Leftrightarrow \tan 2 A = \frac{(\frac{2 \sin Acos A}{\cos^{2} A})}{(\frac{\cos^{2} A - \sin^{2} A}{\cos^{2} A})}

\Leftrightarrow \tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A}

Answered by physicstutes last updated on 23/Jan/21

\tan 2 A = \frac{\sin 2 A}{\cos 2 A} = \frac{2 \sin A \cos A}{\cos^{2} A - \sin^{2} A} = \frac{2 \tan A}{1 - \tan^{2} A}