Trig Proof

**fvvfz** · April 30th 2011, 06:38 PM

So last night I was doing a trig proof and found the solution easily enough but someone else tried to do it a different way and I for the life of me couldn't find any way to do it besides going around in circles to prove.I think it's quite challenging but you might not.

Solve for the left(Work on the right)

tanB+1 = secB+cscB(secB+cscB)
tanB-1 = secB-cscB(secB+cscB)

**HappyJoe** · May 1st 2011, 01:56 AM

That identity doesn't look right to me. For instance, the left hand side is not defined for B = pi/4, because the denominator becomes zero. On the other, the right hand side _is_ defined for B=pi/4.

**Soroban** · May 1st 2011, 05:23 AM

Hello, fvvfz!

I agree with Happy Joe . . . There is a typo.

I believe you are missing parentheses on the right.

$\text{Prove: }\;\frac{\tan\theta + 1}{\tan\theta - 1} \:=\:\frac{(\sec\theta + \csc \theta)(\sec\theta + \csc\theta)}{(\sec\theta - \csc\theta)(\sec\theta + \csc\theta)}$

$\text{Cancel and we have: }\;\frac{\sec\theta + \csc\theta}{\sec\theta - \csc\theta} \;=\;\dfrac{\dfrac{1}{\cos\theta} + \dfrac{1}{\sin\theta}}{\dfrac{1}{\cos\theta} - \dfrac{1}{\sin\theta}}$

$\text{Multiply by }\frac{\sin\theta\cos\theta}{\sin\theta\cos\theta}\!:\;\;\frac{\sin\theta\cos\theta\left(\dfrac{1}{\cos\theta} + \dfrac{1}{\sin\theta}\right)} {\sin\theta\cos\theta \left(\dfrac{1}{\cos\theta} - \dfrac{1}{\sin\theta}\right)} \;=\;\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta}$

$\text{Divide numerator and denominator by }\cos\theta\!:$

. . . . . . $\frac{\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\cos\theta}}{\dfrac{\sin\theta}{\cos\theta} - \dfrac{\cos\theta}{\cos\theta}} \;=\;\frac{\tan\theta + 1}{\tan\theta - 1}$

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