1. ## Verifying!!

1. (cos^2t)/(1-sint)= 1+sint

2. (cot)(x/2)*sinx= 1 + cosx

3. (2cotx)=csc^2x*sin2x

4. (2+2cot^2x)= 2cotxsecxcscx

5. (sinxcosx)=(cosx-cos^3x)/(sinx)

6. 2sinA-cscA=sinA- (cotA)/(secA)

2. Knowing

$\sin^2x+\cos^2x=1$

the number 1:

$
\frac{{\cos ^2 t}}
{{1 - \sin t}} = \frac{{1 - \sin ^2 t}}
{{1 - \sin t}} = \frac{{\left( {1 - \sin t} \right)\left( {1 + \sin t} \right)}}
{{1 - \sin t}} = 1 + \sin t
$

too could be usefull

$
\begin{gathered}
\sin \left( {x + y} \right) = \sin x\cos y + \sin y\cos x \hfill \\
\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y \hfill \\
\end{gathered}
$

3. Hello, J^2!

$2)\;\;\cot\tfrac{x}{2}\cdot\sin x\:=\: 1 + \cos x$

We have: . $\cot\tfrac{x}{2}\;\;\cdot\;\;\sin x$
. . . . . $= \;\overbrace{\frac{\cos\frac{x}{2}}{\sin\frac{x}{2 }}}\cdot\overbrace{2\sin\tfrac{x}{2}\cos\tfrac{x}{ 2}} \;=\;2\cos^2\!\tfrac{x}{2} \;=\;1 + \cos x$

$3)\;\;2\cot x \:=\:\csc^2\!x\cdot\sin2x$

We have: . $\csc^2\!x\;\;\cdot\;\;\sin2x$

. . . . . $= \;\overbrace{\frac{1}{\sin^2\!x}}\cdot\overbrace{2 \sin x\cos x} \;=\;\frac{2\cos x}{\sin x} \;=\;2\cot x$

4. ## THANKS!!

Thank you guys, very much!!! I don't know why I could not figure these ones out... but i guess i am just better at the other 25... Thanks again guys!!