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asdfghjkl8063 Group Title

(tanTheta+secTheta-1)/(tanTheta-secTheta+1) = (1+sinTheta)/(cosTheta) Prove that the two expressions are equal.

  • one year ago
  • one year ago

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  1. vinrar Group Title
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    tanTheta = sinTheta/cosTheta and secTheta=1/cosTheta substitute them

    • one year ago
  2. vinrar Group Title
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    done substituting?

    • one year ago
  3. Snapbacklive Group Title
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    This is an identity. Right side = 0 Left side: = secθ - sinθtanθ - cosθ Change tanθ to sinθ/cosθ secθ - sin^2θ/cosθ - cosθ Multiply the cosθ by cosθ/cosθ and change the secθ to 1/cosθ (1/cosθ) - (sin^2θ/cosθ) - (cos^2θ/cosθ) You now have a common denominator: [1 - sin^2θ - cos^2θ] / [cosθ] Factor out a negative in the numerator [1 - (sin^2θ + cos^2θ)] / [cosθ] sin^2θ + cos^2θ = 1 [1 - (1)] / [cosθ] 0 / [cosθ] 0 So LS = RS, QED.

    • one year ago
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