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Prove that tan(x)sin(x) = sec(x)-cos(x)

tan(x)sin(x) = [sin(x) / cos (x)] sin(x)

= sin2(x) / cos(x)

= [1-cos2(x)] / cos(x)

= 1/cos(x) - cos2(x)/ cos(x)

= sec(x)-cos(x)

Q.E.D

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Q: Sec - cos equals tansin
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Cos plus tansin equals sec?

Start on the left-hand side. cos(x) + tan(x)sin(x) Put tan(x) in terms of sin(x) and cos(x). cos(x) + [sin(x)/cos(x)]sin(x) Multiply. cos(x) + sin2(x)/cos(x) Make the denominators equal. cos2(x)/cos(x) + sin2(x)/cos(x) Add. [cos2(x) + sin2(x)]/cos(x) Use the Pythagorean Theorem to simplify. 1/cos(x) Since 1/cos(x) is the same as sec(x)- the right-hand side- the proof is complete.


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Is cos 2 x sec x equals 2 cos x - sec x an identity?

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Find the principal solution of sec x equals 2?

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