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Question Number 46573 by Rio Michael last updated on 28/Oct/18

$$Showthat$$$$sin2x\equiv \frac{2tanx}{1+{tan}^{2}x}$$

Commented by peter frank last updated on 28/Oct/18

$$\sin 2\mathrm{x}=\frac{2\mathrm{sinxcosx}}{1}$$$$=\frac{2\mathrm{sinxcosx}}{{\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}}$$$$=\frac{\frac{2\mathrm{sinxcosx}}{{\cos }^2\mathrm{x}}}{1+{\tan }^2\mathrm{x}}=\frac{2\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}$$$$\mathrm{please}\mathrm{recheck}$$

Commented by maxmathsup by imad last updated on 28/Oct/18

$$youransweriscorrctpeterbuttheQ.containaerror.$$

Commented by peter frank last updated on 28/Oct/18

$$\mathrm{thank}\mathrm{you}\mathrm{sir}....\mathrm{I}\mathrm{also}\mathrm{had}\mathrm{such}\mathrm{doubt}.$$

Commented by hknkrc46 last updated on 29/Nov/18

$$\tan 2x=\frac{2\tan x}{1-\tan {}^{2}x}\Rightarrow 2\tan x=\tan 2x(1-\tan {}^{2}x)$$$$\{1-\tan {}^{2}x=1-\frac{\sin {}^{2}x}{\cos {}^{2}x}=\frac{\cos {}^{2}x-\sin {}^{2}x}{\cos {}^{2}x}=\frac{\cos 2x}{\cos {}^{2}x}\}$$$$2\tan x=\tan 2x\cdot \frac{\cos 2x}{\cos {}^{2}x}=\frac{\sin 2x}{\cos 2x}\cdot \frac{\cos 2x}{\cos {}^{2}x}=\frac{\sin 2x}{\cos {}^{2}x}$$$$\{1+\tan {}^{2}x=1+\frac{\sin {}^{2}x}{\cos {}^{2}x}=\frac{\sin {}^{2}x+\cos {}^{2}x}{\cos {}^{2}x}=\frac{1}{\cos {}^{2}x}\}$$$$\frac{2\tan x}{1+\tan {}^{2}x}=\frac{\frac{\sin 2x}{\cos {}^{2}x}}{\frac{1}{\cos {}^{2}x}}=\sin 2x$$

Answered by Nabraj Awasthi last updated on 30/Oct/18

$$Leftside$$$$=sin2x=2sinx.cosx$$$$multiplyinganddivingitby{sex}^{2}x,wehave$$$$2sinx.cosx\times \frac{{sec}^{2}x}{{sec}^{2}x}=\frac{2sinx.secx}{{sec}^{2}x}=\frac{2tanx}{1+{tan}^{2}x}$$

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