A trigonemetry problem If \(\displaystyle{\sec{{\left(\theta+\alpha\right)}}}{\sec{{\left(\theta-\alpha\right)}}}={2}{\sec{{\left(\theta\right)}}}\) and \(\displaystyle{\cos{{\left(\theta\right)}}}={K}{\cos{{\frac{{\alpha}}{{{2}}}}}}\) Then prove

Anika Wolfe

Anika Wolfe

Answered question

2022-03-16

A trigonemetry problem
If sec(θ+α)sec(θα)=2sec(θ)
and cos(θ)=Kcosα2
Then prove that K=±2
I have converted the sec into cos and then applied the formulaes of cos(θ+α)cos(θ+α) we get
2sin2α=2cos2θcosθ
Then 2cos2(θ)=K2(1+cosα) from the other equation
But i can't get K by solving both the equations

Answer & Explanation

bridgedcuu

bridgedcuu

Beginner2022-03-17Added 7 answers

First solve sec(θ+α)+sec(θα)=2sec(θ)
and then compare it to cos(θ)=Kcos{α2}
We have
sec(θ+α)+sec(θα)=2sec(θ)
1cos(θ+α)+1cos(θα)=2cosθ
cos(θ+α)+cos(θα)cos(θ+α)cos(θα)=2cos(θ)
2cosθcosαcos2θ+cos2α=1cosθ
2cos2θcosα=cos2θ+cos2α
2cos2θ(cosα1)=2cos2α2
cos2θ=cos2α1cosα1=cosα+1
cos2θ=2cos2(α2)
cosθ=±2cos(α2)
When we compare this to cos(θ)=Kcos{(α2)} we get
K=±2

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