Help Me With This Trig ID proof Please Soon!

**chiro** · November 3rd 2012, 04:05 PM

Hey Melcarthus.

Your equation is wrong. As an example let theta = 0, then on the LHS you get 0/0 for the answer but on the RHS you get 1/1.

The equation is screwed up.

**Soroban** · November 3rd 2012, 07:42 PM

Hello, Melcarthus!

$\text{Prove: }\:\frac{\sin\theta -\cos\theta + 1}{\sin\theta + \cos\theta - 1} \:=\:\frac{\sin\theta + 1}{\cos\theta}$

The left side is: . $\frac{\sin\theta - (\cos\theta - 1)}{\sin\theta + (\cos\theta - 1)}$

Then: . $\frac{\sin\theta - (\cos\theta-1)}{\sin\theta + (\cos\theta - 1)} \cdot \color{blue}{\frac{\sin\theta - (\cos\theta - 1)}{\sin\theta - (\cos\theta -1)}}$

. . . . . $=\;\frac{\sin^2\!\theta - 2\sin\theta(\cos\theta-1) + (\cos\theta-1)^2}{\sin^2\!\theta - (\cos\theta -1)^2}$

. . . . . $=\;\frac{\sin^2\!\theta - 2\sin\theta\cos\theta + 2\sin\theta + \cos^2\!\theta - 2\cos\tjeta + 1}{\sin^2\!\theta - \cos^2\!\theta + 2\cos\theta - 1}$

. . . . . $=\;\frac{\overbrace{\sin^2\!\theta + \cos^2\!\theta}^{\text{This is 1}} + 1 + 2\sin\theta - 2\cos\theta - 2\sin\theta\cos\theta}{2\cos\theta - \cos^2\theta - \underbrace{(1-\sin^2\!\theta)}_{\text{This is }\cos^2\theta}}$

. . . . . $=\;\frac{2 + 2\sin\theta - 2\cos\theta - 2\sin\theta\cos\theta}{2\cos\theta - 2\cos^2\!\theta}$

. . . . . $=\;\frac{2(1+\sin\theta) - 2\cos\theta(1 + \sin\theta)}{2\cos\theta(1-\cos\theta)}$

. . . . . $=\;\frac{2(1+\sin\theta)(1-\cos\theta)}{2\cos\theta(1-\cos\theta)}$

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

**StefanTM** · November 5th 2012, 08:20 AM

Hi Melcartus,
a alternative solution:
(sint-cost+1)*cost=(sint+cost-1)*(sint+1)
You receive a identity, not forget, that sin^2(t)+cos^2(t)=1.

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