Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 111267 by I want to learn more last updated on 03/Sep/20

Answered by Her_Majesty last updated on 03/Sep/20

$$tan\alpha -tan\beta =\frac{sin(\alpha -\beta )}{cos\alpha cos\beta };cos\alpha cos\beta =\frac{cos(\alpha -\beta )+cos(\alpha +\beta )}{2}$$$$\frac{a}{b}=\frac{tan10°}{tan50°-tan40°}=\frac{\frac{sin10°}{cos10°}}{\frac{sin10°}{cos50°cos40°}}=\frac{cos50°cos40°}{cos10°}=$$$$=\frac{\frac{cos10°+cos90°}{2}}{cos10°}=\frac{\frac{cos10°}{2}}{cos10°}=\frac{1}{2}$$

Commented by I want to learn more last updated on 03/Sep/20

$$\mathrm{Thanks}\mathrm{ma},\mathrm{i}\mathrm{appreciate}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com