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Question Number 78311 by mathocean1 last updated on 15/Jan/20

$$\mathrm{Show}\mathrm{that}\frac{3+\sin 2x-2\cos 2x}{1+3{sin}^{2}x-cos2x}=\frac{2}{5}(2+\frac{1}{tanx})$$

Commented by jagoll last updated on 16/Jan/20

$$\frac{3+2\sin x\cos x-2(1-2\sin {}^{2}x)}{1+3\sin {}^{2}x-(1-2\sin {}^{2}x)}=$$$$\frac{1+2\sin x\cos x+4\sin {}^{2}x}{5\sin {}^{2}x}=$$$$\frac{1}{5}\mathrm{cosec}{}^{2}x+\frac{2}{5}\cot x+\frac{4}{5}=$$$$\frac{1}{5}\{\mathrm{cosec}{}^{2}x+2\cot x+4\}$$

Commented by john santu last updated on 16/Jan/20

$$\frac{1}{5}\{1+\cot {}^{2}x+2\cot x+4\}=$$$$\frac{1}{5}\{\frac{1}{\tan {}^{2}x}+\frac{2}{\tan x}+5\}=$$$$1+\frac{1}{5\tan {}^{2}x}+\frac{2}{5\tan x}$$

Answered by mind is power last updated on 15/Jan/20

$$\cos \left(2\mathrm{x}\right)=-{2\sin }^2\left(\mathrm{x}\right)+1\sin \left(2\mathrm{x}\right)=2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)\Rightarrow$$$$\frac{3+\sin \left(2\mathrm{x}\right)-2\cos \left(2\mathrm{x}\right)}{1+{3\sin }^2\left(\mathrm{x}\right)-\cos \left(2\mathrm{x}\right)}=\frac{3+2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)+{4\sin }^2\left(\mathrm{x}\right)-2}{1+{3\sin }^2\left(\mathrm{x}\right)+{2\sin }^2\left(\mathrm{x}\right)-1}$$$$=\frac{{4\sin }^2\left(\mathrm{x}\right)+2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)+1}{{5\sin }^2\left(\mathrm{x}\right)}$$$$=\frac{4}{5}+\frac{2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)}{{5\sin }^2\left(\mathrm{x}\right)}+\frac{1}{{5\sin }^2\left(\mathrm{x}\right)}$$$$=\frac{4}{5}+\frac{2\cos \left(\mathrm{x}\right)}{5\sin \left(\mathrm{x}\right)}+\frac{1}{{5\sin }^2\left(\mathrm{x}\right)}=\frac{4}{5}+\frac{2}{5\tan \left(\mathrm{x}\right)}+\frac{{\sin }^2\left(\mathrm{x}\right)+{\cos }^2\left(\mathrm{x}\right)}{{5\sin }^2\left(\mathrm{x}\right)}$$$$=\frac{4}{5}+\frac{2}{5\tan \left(\mathrm{x}\right)}+\frac{1}{5}+\frac{1}{{5\tan }^2\left(\mathrm{x}\right)}=1+\frac{2}{5\tan \left(\mathrm{x}\right)}+\frac{1}{{5\tan }^2\left(\mathrm{x}\right)}$$$$$$$$$$

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