Prove sinx=cos(90-x) and tanx=1/tan(90-x) in geometry's way,and prove tanx=sinx/cosx in algebra's way, thank you so much!!!^_^

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Consider right triangle `Delta ABC` with acute angles A and B. Note that A and B are complements; that is the measure of angle A equals 90 minus the measure of angle B.

Now by right triangle trigonometry the sine of an angle is defined to be the ratio of the leg opposite the angle to the hypotenuse, while the definition of the cosine of an angle is the ratio of the leg adjacent to the angle to the hypotenuse.

Thus `sinA=a/c` and `cosB=a/c` . But B=90-A so `sinA=cos(90-A)` as required.

The tangent of an angle is defined as the leg opposite the angle to the leg adjacent to the angle. `tanA=a/b,tanB=b/a` . Then `tanA=1/(tan(B))=1/(tan(90-A))` as required.

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Show `tanx=(sinx)/(cosx)` :

Using the set-up above:

`tanA=a/b=((a/c))/((b/c))=(sinA)/(cosA)`

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