If secθ + tanθ = p, …
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Yogita Ingle 6 years, 10 months ago
Given, sec θ + tan θ = p ...........1
We know that
sec2 θ - tan2 θ = 1
=> (sec θ - tan θ)(sec θ + tan θ) = 1
=> (sec θ - tan θ) * p = 1
=> sec θ - tan θ = 1/p ............2
Add equation 1 and 2, we get
2 * sec θ = p + 1/p
=> 2 * sec θ = (p2 + 1)/p
=> sec θ = (p2 + 1)/2p
and cos θ = 1/sec θ
=> cos θ = 2p/(p2 + 1)
Subtract equation 1 and 2, we get
2 * tan θ = p - 1/p
=> 2 * tan θ = (p2 - 1)/p
=> tan θ = (p2 - 1)/2p
and cot θ = 1/tan θ
=> cot θ = 2p/(p2 - 1)
We know that
tan θ = sin θ/ cos θ
=> sin θ = tan θ * cos θ
=> sin θ = {(p2 - 1)/2p} * {2p/(p2 + 1)}
=> sin θ = (p2 - 1)/(p2 + 1)
and cosec θ = 1/sin θ
=> cosec θ = (p2 + 1)/(p2 - 1)
2Thank You