I'm having a hard time understanding how to verify trig functions using trig identities.

can someone walk me through how to verify this equation? thanks

(cscθ)(tanθ) = secθ

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- Feb 6th 2011, 11:59 AMTacticalProHelp me with trig functions!!
I'm having a hard time understanding how to verify trig functions using trig identities.

can someone walk me through how to verify this equation? thanks

(cscθ)(tanθ) = secθ - Feb 6th 2011, 12:04 PMskeeter
- Feb 6th 2011, 12:06 PMhomeylova223
It is like this

(1/sinx)(sinx/cosx)= sec

(1/cosx)= secx

secx=secx - Feb 6th 2011, 12:07 PMpickslides
I can help

$\displaystyle \displaystyle \csc \theta \tan \theta$

$\displaystyle \displaystyle = \frac{1}{\sin \theta}\times \frac{\sin \theta}{\cos \theta}$

$\displaystyle \displaystyle =\frac{1}{\cos \theta}$

$\displaystyle \displaystyle =\sec\theta$ - Feb 6th 2011, 12:26 PMTacticalPro
okay thanks! what about this problem?

csc[(pi/2)-θ] / tan(-θ) = -cscθ

i reduced that down to

secθ / tanθ = cscθ

and i'm stuck. can anyone help me with what I do next? thanks - Feb 6th 2011, 12:32 PMArchie Meade
- Feb 6th 2011, 12:45 PMTacticalPro
- Feb 6th 2011, 01:02 PMArchie Meade
No, I didn't forget.

The minus sign only comes at the last step.

I'm working from the left and evaluating that.

Now, what is $\displaystyle sec\theta cos\theta\;\;?$ - Feb 6th 2011, 02:19 PMTacticalPro
- Feb 6th 2011, 02:21 PMdwsmith
- Feb 6th 2011, 02:24 PMTacticalPro
ok so now i'm left with:

1/-sinθ = -cscθ

cscθ = cscθ - Feb 6th 2011, 02:26 PMdwsmith
- Feb 6th 2011, 03:07 PMArchie Meade
You're probably overanalysing the two sides being equal.

You only need evaluate the left hand side to see what it eventually yields...

$\displaystyle \displaystyle\frac{1}{-sin\theta}=-\frac{1}{sin\theta}=-csc\theta$