Question

If u know plz answer it and if satisfactory I will mark it as brainliest

Answer 1

Taking tetha as A

CosA2 / CotA2-CosA2 = 3

CosA2 = 3 ( CotA2-CosA2)
CosA2 = 3CotA2-3CosA2
4CosA2 = 3CotA2
4CosA2 = 3(CosA2/SinA2)
4CosA2 = 3CosA2/SinA2
4/3 CosA2/CosA2 = 1/SinA2. { CosA2
Taking Reciprocals..
3/4 = SinA2
SinA = √3/4
SinA = √3/2
SinA = Sin 60°
A = 60°
Hence tetha is 60°

VERIFICATION:-

CosA2 / CotA2 - CosA2
Cos60^2 / Cot60^2 - Cos60^2
(1/2)^2 / (1/√3)^2 - (1/2)^2
= 1/4 ÷ 1/3-1/4
= 1/4 ÷ 4-3/12 { Taking LCM }
= 1/4÷1/12
= 12/4 = 3
Hence Verified!


Hope this helps!! cheers!! (:

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