How do you prove $(sec (θ)) + (tan (θ)) (1 - sin (θ)) = cos (θ)$ $(sec(theta))+(tan(theta))(1-sin(theta))=cos(theta)$ ?

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Answer 1 · 2018-05-24T14:45:41

For a Proof, please refer to Explanation.

$(sec θ + tan θ) (1 - sin θ)$ $(sectheta+tantheta)(1-sintheta)$ ,

$= (\frac{1}{cos θ} + \frac{sin θ}{cos θ}) (1 - sin θ)$ $=(1/costheta+sintheta/costheta)(1-sintheta)$ ,

$= (\frac{1 + sin θ}{cos θ}) (\frac{1 - sin θ}{1})$ $=((1+sintheta)/costheta)((1-sintheta)/1)$ ,

$= \frac{1 - {sin}^{2} θ}{cos θ}$ $=(1-sin^2theta)/costheta$ ,

$= \frac{{cos}^{2} θ}{cos θ}$ $=cos^2theta/costheta$ ,

$= cos θ$ $=costheta$ , as desired!

How do you prove (sec(theta))+(tan(theta))(1-sin(theta))=cos(theta)(sec(θ))+(tan(θ))(1−sin(θ))=cos(θ)(sec(theta))+(tan(theta))(1-sin(theta))=cos(theta)?