Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 117259 by bemath last updated on 10/Oct/20

Answered by Dwaipayan Shikari last updated on 10/Oct/20

tanα+tanβ=3  tanαtanβ=−3  tan(α+β)=((tanα+tanβ)/(1−tanαtanβ))=(3/4)  sin(α+β)=(3/5) ,cos(α+β)=(4/5)  ∣sin^2 (α+β)−3sin(α+β)cos(α+β)−3cos^2 (α+β)∣  =∣(9/(25))−((36)/(25))−((48)/(25))∣=∣−((75)/(25))∣=3

tanα+tanβ=3tanαtanβ=3tan(α+β)=tanα+tanβ1tanαtanβ=34sin(α+β)=35,cos(α+β)=45sin2(α+β)3sin(α+β)cos(α+β)3cos2(α+β)=∣92536254825∣=∣7525∣=3

Commented by bemath last updated on 10/Oct/20

santuyyy

santuyyy

Answered by bemath last updated on 10/Oct/20

Terms of Service

Privacy Policy

Contact: info@tinkutara.com