prove the identity

[icelated]

I am having a bit of trouble with proving these 2 identities. Normally, i can figure these out. there is no solution in the back of the book so here i am.

first identity

1 - cos(2t) / cos(t)sin(t) = 2 tan(t)

what i have come up with is this. which i know is totally wrong.

1 - cos(2t) / cos(t)sin(t) = 2 tan(t)

= 1 - 2 cos^2(t) - 1 / cos(t) sin(t)  // because cos(2t) = 2cos^2(theta) -1 ???

= Now, i am not sure what to do because for it to be tan it needs to be sint/ cost 
and thats not possible! =(

the next identity

1 / 1 - sin(t) + 1 / 1 + sin(t) = 2sec^2(t)

1 / 1 - sin(t) + 1 / 1 + sin(t) = 2sec^2(t)

use conjugate?

= 1 + sin(t) / 1 - sin(t) + 1 - sin(t) / 1 + sin(t)

= 2 - sin(t) / 1+ sin(t) - sin(t) - sin(t)  // distributing the bottom

thats all i have, and i dont think its correct! i know sec^2(t) = tan^2(t) + 1 but i dont know if that plays any role here?

Any help is much appreciated because i have tried everything i could think of and i cant solve these two problems.

 

[BigGlenntheHeavy]

$\displaystyle For \ the \ first \ problem \ use \ the \ identity \ cos(2t) \ = \ 1-2sin^2(t).$

 

[soroban]

Hello, icelated!

Here's the second one . . .

$\displaystyle \frac{1}{1 - \sin t} + \frac{1}{1 + \sin t} \:=\: 2\sec^2\!t$

On the left side, add the two fractions:

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

 

[icelated]

Thank you so much guys. How easy they are now! =)
do i do anything to this post since its what i needed?