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Question Number 117259 by bemath last updated on 10/Oct/20
Answered by Dwaipayan Shikari last updated on 10/Oct/20
tanα+tanβ=3tanαtanβ=−3tan(α+β)=tanα+tanβ1−tanαtanβ=34sin(α+β)=35,cos(α+β)=45∣sin2(α+β)−3sin(α+β)cos(α+β)−3cos2(α+β)∣=∣925−3625−4825∣=∣−7525∣=3
Commented by bemath last updated on 10/Oct/20
santuyyy
Answered by bemath last updated on 10/Oct/20
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