If Sin =a^2-b^2\a^2+b^2 , find the …
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Posted by Rajat Arora 4 years, 8 months ago
- 2 answers
Gaurav Seth 4 years, 8 months ago
In ΔABC, Let ∠ABC be θ
Sin θ = (a2 - b2) / (a2 + b2)
⇒ AB = (a2 - b2)
⇒ AC = (a2 + b2)
⇒ BC = √[(a2 + b2)2 - (a2 - b2)2] [According to Pythagoras theorem]
⇒ BC = √(4a2b2)
⇒ BC = 2ab
Cos θ = 2ab / (a2 + b2)
Tan θ = (a2 - b2) / 2ab.
Cosec θ, Sec θ and Cot θ are the reciprocals of Sin θ, Cos θ, Tan θ respectively.
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Rajat Arora 4 years, 8 months ago
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