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Question Number 113151 by ZiYangLee last updated on 11/Sep/20

$$\mathrm{Prove}\frac{\sin 2\mathrm{A}+\cos 2\mathrm{A}+1}{\sin 2\mathrm{A}+\cos 2\mathrm{A}-1}=\frac{\tan (\mathrm{A}+45°)}{\mathrm{tanA}}$$$$\mathrm{Hence},\mathrm{prove}\mathrm{that}\tan 15°=2-\sqrt{3}$$

Commented by ZiYangLee last updated on 11/Sep/20

$$\mathrm{how}\mathrm{to}\mathrm{prove}\tan 15=2-\sqrt{3}??$$

Commented by Dwaipayan Shikari last updated on 11/Sep/20

$$tan\frac{\pi }{12}=\frac{sin\frac{\pi }{12}}{cos\frac{\pi }{12}}=\frac{2{sin}^2\frac{\pi }{12}}{2sin\frac{\pi }{12}cos\frac{\pi }{12}}=\frac{1-cos\frac{\pi }{6}}{sin\frac{\pi }{6}}=\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=2-\sqrt{3}$$

Commented by Dwaipayan Shikari last updated on 11/Sep/20

$$A=-\frac{\pi }{6}$$

Commented by ZiYangLee last updated on 11/Sep/20

$$\mathrm{thanks}!!\mathrm{i}\mathrm{get}\mathrm{it}!!<$$

Answered by som(math1967) last updated on 11/Sep/20

$$\frac{1+\sin 2\mathrm{A}+\cos 2\mathrm{A}}{\cos 2\mathrm{A}-(1-\sin 2\mathrm{A})}$$$$=\frac{{(\mathrm{cosA}+\mathrm{sinA})}^2+(\mathrm{cosA}+\mathrm{sinA})(\mathrm{cosA}-\mathrm{sinA})}{(\mathrm{cosA}+\mathrm{sinA})(\mathrm{cosA}-\mathrm{sinA})-{(\mathrm{cosA}-\mathrm{sinA})}^2}$$$$=\frac{(\mathrm{cosA}+\mathrm{sinA})(\mathrm{cosA}+\mathrm{sinA}+\mathrm{cosA}-\mathrm{sinA})}{(\mathrm{cosA}-\mathrm{sinA})(\mathrm{cosA}-\mathrm{sinA}-\mathrm{cosA}+\mathrm{sinA})}$$$$\frac{(\mathrm{cosA}+\mathrm{sinA})2\mathrm{cosA}}{(\mathrm{cosA}-\mathrm{sinA})2\mathrm{sinA}}$$$$=\frac{\frac{\mathrm{cosA}+\mathrm{sinA}}{\mathrm{cosA}}}{\frac{\mathrm{cosA}-\mathrm{sinA}}{\mathrm{cosA}}}\times \mathrm{cotA}$$$$=\frac{1+\mathrm{tanA}}{1-\mathrm{tanA}}\times \frac{1}{\mathrm{tanA}}$$$$=\frac{\tan (\mathrm{A}+45°)}{\mathrm{tanA}}★$$$$★\frac{1+\mathrm{tanA}}{1-\mathrm{tanA}}=\tan (\mathrm{A}+45°)$$$$$$

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