answersLogoWhite

0

Let: o = opposite

h = hypotenuse

a = adjacent

sin = o/h; tan = o/a

Therefore, sin/tan = (o/h)/(o/a)

= (o/h)*(a/o)

= a/h

= cos

User Avatar

Wiki User

13y ago

Still curious? Ask our experts.

Chat with our AI personalities

CoachCoach
Success isn't just about winning—it's about vision, patience, and playing the long game.
Chat with Coach
ReneRene
Change my mind. I dare you.
Chat with Rene
BlakeBlake
As your older brother, I've been where you are—maybe not exactly, but close enough.
Chat with Blake

Add your answer:

Earn +20 pts
Q: How do you show that sinx divided by tanx equals cosx?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Continue Learning about Math & Arithmetic

How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?

(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx


Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?

2


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

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.


How do you prove the following equation the quantity of sin theta divided by 1 minus cos theta minus the quantity 1 plus cos theta divided by sin theta equals 0?

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


Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?

(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx