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Prove that \displaystyle \frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta}{\cos \theta}

​1+sinθ

​cosθ
​​ =
​cosθ

​1−sinθ
​​

​​

Prove that \displaystyle \frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta-example-1

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Prove that \displaystyle \frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta-example-1
User Muhammad Mahmoud
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7.3k points
2 votes

Answer:

sin^2(θ)+cos^2(θ)=1

Explanation:

We know that the statement above is true because of the Pythagorean identity theorem, which states the aforementioned equation. If you solve the equation for 1 you get the same equation.

To do this first multiply both sides by cos(θ), this gives you (cos^2θ)/1+sinθ = 1-sinθ

Then, multiply both sides by sinθ. This equals cos^θ=1-sin^2θ.

Finally, add sin^2θ to both sides. This equals the final answer of cos^2θ+sin^2θ=1. Which is true.

Prove that \displaystyle \frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta-example-1
User Billy Kimble
by
6.5k points