## ianwalters Group Title verify: secx - cosx = tanxsinx one year ago one year ago

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1. whpalmer4 Group Title

What's the definition of $$\sec x$$?

2. ianwalters Group Title

cosx

3. whpalmer4 Group Title

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

4. ianwalters Group Title

oh oops. its 1/cosx

5. whpalmer4 Group Title

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

6. ianwalters Group Title

1/cosx - cosx = tanxsinx

7. whpalmer4 Group Title

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

8. ianwalters Group Title

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

9. whpalmer4 Group Title

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

10. ianwalters Group Title

yes but i cant remember what that is equivalent to

11. whpalmer4 Group Title

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 $$\sin^2x = 1-\cos^2x$$

12. whpalmer4 Group Title

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

13. whpalmer4 Group Title

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!

14. ianwalters Group Title

thank you so much!