Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 37663 by Rio Mike last updated on 16/Jun/18

$$\mathrm{show}\mathrm{that}\frac{sin2A}{1+cos2A}=\mathrm{tanA}.$$

Answered by Joel579 last updated on 16/Jun/18

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

Answered by kunal1234523 last updated on 16/Jun/18

$$\frac{2\sin A\cos A}{({\sin }^{2}A+{\cos }^{2}A)+({\cos }^{2}A-{\sin }^{2}A)}$$$$=\frac{2\sin A}{2\cos A}$$$$=\tan A$$

Answered by Ahmed Neutron last updated on 17/Jun/18

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com