1. ## Verification problems

1. sinx+cosx/tan^2x-1 = cos^2x/sinx-cosx

I've tried everything(converting to sines and cosines, factoring, etc.) and even my teacher couldn't figure it out...

2. 1+tanx/1-tanx = sec^2x+2tanx/2-sec^2x

I make everything more and more complicated when I try and verify this one from the right side

2. Originally Posted by clay10
1. sinx+cosx/tan^2x-1 = cos^2x/sinx-cosx

I've tried everything(converting to sines and cosines, factoring, etc.) and even my teacher couldn't figure it out...

2. 1+tanx/1-tanx = sec^2x+2tanx/2-sec^2x

I make everything more and more complicated when I try and verify this one from the right side
$\displaystyle \frac{\sin{x} + \cos{x}}{\tan^2{x} - 1}$

multiply numerator and denominator by $\displaystyle \cos^2{x}$ ...

$\displaystyle \frac{\cos^2{x}\sin{x} + \cos^3{x}}{\sin^2{x} - \cos^2{x}}$

factor ...

$\displaystyle \frac{\cos^2{x}(\sin{x} + \cos{x})}{(\sin{x} - \cos{x})(\sin{x}+\cos{x})}$

$\displaystyle \frac{\cos^2{x}}{\sin{x} - \cos{x}}$

for the second one, change $\displaystyle sec^2{x}$ to $\displaystyle \tan^2{x} + 1$ on the RHS, then factor both numerator and denominator.

3. thanks a lot. I was confused why you chose cos^2x but I realized we have never used any identities from the other side of the equation to the more complicated side.