cosec theta+cot theta=√3,sin theta=?
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According to the problem:
Cosec θ + Cot θ = √3 -----(1)
We know that ,
Cosec² θ - Cot² θ = 1
( Cosec θ + Cot θ )(Cosec θ - Cot θ) =1
[ Since a² - b² = ( a + b ) ( a - b ) ]
( √3 ) ( Cosec θ - Cot θ ) = 1 [from (1) ]
Cosec θ - Cot θ = 1 / √3 -----(2)
On adding equations (1) and (2) we get,
2Cosec θ = √3 + 1 /√3
⇒ 2Cosec θ = ( 3 + 1 ) / √3
⇒ 2Cosec θ = 4 / √3
⇒ Cosec θ = 4 /2√3
⇒ Cosec θ = 2 / √3
⇒ Cosec θ = Cosec 60°
∴ θ = 60°
=> Sin θ = sin 60°
=> √3 / 2
Hope it helps :)
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