Question

verify:
secx - cosx = tanxsinx

Answer 1

What's the definition of \(\sec x\)?

Answer 2

cosx

Answer 3

No. Why would we need a different name if it is the same?

Answer 4

oh oops. its 1/cosx

Answer 5

There we go. So what if you substitute 1/cos x in to the equation, what do you get?

Answer 6

1/cosx - cosx = tanxsinx

Answer 7

Okay, can you make a common denominator and combine 1/cos x - cos x?

Answer 8

now: (1-cos^2x)/cosx

Answer 9

okay, does that numerator look like any of your trig identities?

Answer 10

yes but i cant remember what that is equivalent to

Answer 11

well, the formula for a circle at the origin is \(x^2 + y^2 = r^2\) or if we are talking the unit circle used to define the trig functions, \x^2 + y^2 = 1\) and so \(\sin^2x + \cos^x = 1\) or \(\sin^2x = 1-\cos^2x\)

Answer 12

rats, left out a '('!
\(x^2+y^2=1\)

Answer 13

So our left hand side is now \[\frac{\sin^2x}{\cos x}\]Wouldn't it be great if \[\tan x = \frac{\sin x}{\cos x}\] because that would make our right hand side be \[\frac{\sin x}{\cos x}*\sin x = \frac{\sin^2x}{\cos x}\]just like our left hand side.

SOH CAH TOA
sine = opposite over hypotenuse
cosine = adjacent over hypotenuse
tangent = opposite over adjacent

Hey, look at that: opposite over hypotenuse divided by adjacent over hypotenuse = opposite over adjacent, so \[\tan x =\frac{\sin x}{\cos x}\]just like we want!

Answer 14

thank you so much!