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Question

Vocaloid · Answer

thoughts? recall the double angle cos identities (here's a reference: https://www2.clarku.edu/faculty/djoyce/trig/doublecosine.jpg) notice what happens if you replace cos(2x) in the numerator with cos^2(x) - sin^2(x) $$\frac{ \cos(2x) }{ \cos(x) }=\frac{ \cos^{2}(x)-\sin^{2}(x) }{ \cos(x) }=\frac{ \cos^{2}(x) }{ \cos(x) }-\frac{ \sin^{2}(x) }{ \cos(x) }=\cos(x)-\sin(x)\tan(x)$$ (so identity I is true). you can prove one of the other identities if you replace cos(2x) with 1-2sin^2(x) and simplifying in a similar manner.

kekeman · Answer

@vocaloid wrote:

thoughts? recall the double angle cos identities (here's a reference: https://www2.clarku.edu/faculty/djoyce/trig/doublecosine.jpg) notice what happens if you replace cos(2x) in the numerator with cos^2(x) - sin^2(x) $$\frac{ \cos(2x) }{ \cos(x) }=\frac{ \cos^{2}(x)-\sin^{2}(x) }{ \cos(x) }=\frac{ \cos^{2}(x) }{ \cos(x) }-\frac{ \sin^{2}(x) }{ \cos(x) }=\cos(x)-\sin(x)\tan(x)$$ (so identity I is true). you can prove one of the other identities if you replace cos(2x) with 1-2sin^2(x) and simplifying in a similar manner.

I feel that the answer is option 3: I and II

Vocaloid · Answer

That’s correct, substituting 1 - 2sin^2(x) will result in identity II so I and I are true. III is not equal to the original expression.