Best Answer

From the Pythagorean identity, sin2x = 1-cos2x.

LHS = 1/(sinx cosx) - cosx/sinx

LHS = 1/(sinx cosx) - (cosx/sinx)(cosx/cosx)

LHS = 1/(sinx cosx) - cos2x/(sinx cosx)

LHS = (1- cos2x)/(sinx cosx)

LHS = sin2x /(sinx cosx) [from Pythagorean identity]

LHS = sin2x /(sinx cosx)

LHS = sinx/cosx

LHS = tanx [by definition]

RHS = tanx

LHS = RHS and so the identity is proven.

Q.E.D.

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Q: Can you Show 1 over sinx cosx - cosx over sinx equals tanx?

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You will have to bear with the angle being represented by x because this browser will not allow characters from other alphabets!sin^2x + cos^2x = 1=> sin^2x = 1 - cos^x = (1 + cosx)(1 - cosx)Divide both sides by sinx (assuming that sinx is not zero).=> sinx = (1 + cosx)(1 - cosx)/sinxDivide both sides by (1 - cosx)=> sinx/(1 - cosx) = (1 + cosx)/sinx=> sinx/(1 - cosx) - (1 + cosx)/sinx = 0

cscx-sinx=(cosx)(cotx) 1/sinx-sinx=(cosx)(cosx/sinx) (1/sinx)-(sin^2x/sinx)=cos^2x/sinx cos^2x/sinx=cos^2x/sinx Therefore LS=RS You have to remember some trig identities when answering these questions. In this case, you need to recall that sin^2x+cos^2x=1. Also, always switch tanx cotx cscx secx in terms of sinx and cosx.

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