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

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

Answered by EDWIN88 last updated on 23/Jan/21

$$\mathrm{by}{\mathrm{De}}^{\prime }\mathrm{Moivre}\mathrm{theorem}$$$$\cos 2\mathrm{A}+i\sin 2A={(\cos \mathrm{A}+i\sin \mathrm{A})}^2$$$$=({\cos }^2\mathrm{A}-\sin {}^2\mathrm{A})+i\left(2\sin \mathrm{Acos}\mathrm{A}\right)$$$$\frac{\cos 2\mathrm{A}+i\sin 2\mathrm{A}}{\cos 2\mathrm{A}}=\frac{\cos {}^2\mathrm{A}-\sin {}^2\mathrm{A}}{\cos 2\mathrm{A}}+i\frac{2\sin \mathrm{Acos}\mathrm{A}}{\cos 2\mathrm{A}}$$$$1+i\tan 2\mathrm{A}=1+i\frac{2\sin \mathrm{Acos}\mathrm{A}}{\cos {}^2\mathrm{A}-\sin {}^2\mathrm{A}}$$$$\iff \tan 2\mathrm{A}=\frac{\left(\frac{2\sin \mathrm{Acos}\mathrm{A}}{\cos {}^2\mathrm{A}}\right)}{\left(\frac{\cos {}^2\mathrm{A}-\sin {}^2\mathrm{A}}{\cos {}^2\mathrm{A}}\right)}$$$$\iff \tan 2\mathrm{A}=\frac{2\tan \mathrm{A}}{1-\tan {}^2\mathrm{A}}$$$$$$

Answered by physicstutes last updated on 23/Jan/21

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

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