Proving Identities

[Guest]

Prove that this equation is an identity.
0 = Delta Sign

(sec 0 - tan 0)^2 + 1
------------------------ = 2 tan 0
csc 0(sec 0 - tan 0)

 

[stapel]

It is usually helpful to convert everything to sines and cosines, and see where that leads. So start with the left-hand side, and see what you get.

(Or if you have started working this exercise a different way, please reply showing those steps, instead.)

Thank you.

Eliz.

 

[Guest]

I tried doing that but I get stuck after converting all non-sin/cos into sin/cos.

 

[skeeter]

[(secx - tanx)^2 + 1]/[cscx(secx - tanx)] =

[sec^2(x) - 2secxtanx + tan^2(x) + 1]/[cscx(secx - tanx)] =

{note: tan^2(x) + 1 = sec^2(x), right? ... }

[2sec^2(x) - 2secxtanx]/[cscx(secx - tanx)] =

2secx(secx - tanx)/[cscx(secx - tanx)] =

you can finish up from here

 

[Guest]

I'm so lost atm.

 

[soroban]

Hello, BrianU!

We can do this one head-on if we know that: $\displaystyle \,\tan^2\theta\,+\,1\:=\:\sec^2\theta$

\(\displaystyle \text{Prove: }\L\frac{(\sec\theta\,-\tan\theta)^2\,+\,1}{\csc\theta(\sec\theta\,-\,\tan\theta}\)$\displaystyle \;=\;2\cdot\tan\theta$

The left side is: \(\displaystyle \L\,\frac{\sec^2\theta\,-\,2\cdot\sec\theta\cdot\tan\theta \,+\,\overbrace{\tan^2\theta\,+\,1}}{\csc\theta(\sec\theta\,-\,\tan\theta)}\;=\;\frac{\sec^2\theta\,-\,2\cdot\sec\theta\cdot\tan\theta\,+\,\sec^2\theta}{\csc\theta(\sec\theta\,-\,\tan\theta)}\)

\(\displaystyle \L\;\;\;= \;\frac{2\cdot\sec^2\theta\,-\,2\sec\theta\cdot\tan\theta}{\csc\theta(\sec\theta\,-\,\tan\theta)} \;=\;\frac{2\cdot\sec\theta(\sec\theta\,-\,\tan\theta)}{\csc\theta(\sec\theta\,-\,\tan\theta)} \;= \;\frac{2\cdot\sec\theta}{\csc\theta}\)

\(\displaystyle \L\:\;\;\;= \;\frac{2\cdot\frac{1}{\cos\theta}}{\frac{1}{\sin\theta}} \;= \;2\cdot\frac{\sin\theta}{\cos\theta}\)$\displaystyle \;=\;2\cdot\tan\theta\;\;$ . . . ta-DAA!

 

[Guest]

Thanks a bunch!!!