1. ## Proving Problems

Have an assignment due in the near future, and these four are giving me a hassle to prove, any help or advice would be appreciated.

Prove
$
\frac{1+tan{x}}{sinx}= cscxsecx
$

$
tanx+cotx=\frac{csc^2{x}}{cotx}
$

2. Originally Posted by nightrider456
Have an assignment due in the near future, and these four are giving me a hassle to prove, any help or advice would be appreciated.

Prove
$
\frac{1+sin{x}}{sinx}= cscxsecx
$

$
tanx+cotx=\frac{csc^2{x}}{cotx}
$

$
\frac{Cos2{x}}{sinx}= \frac{Cot^2{x}-1}{csc{x}}
$

$
Sin2x(tanx+cotx)=2
$
$\tan{x} + \cot{x} = \frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}$

$= \frac{\sin^2{x}}{\sin{x}\cos{x}} + \frac{\cos^2{x}}{\sin{x}\cos{x}}$

$= \frac{1}{\sin{x}\cos{x}}$

$= \frac{\sin{x}}{\sin^2{x}\cos{x}}$

$= \frac{1}{\sin^2{x}}\cdot \frac{\sin{x}}{\cos{x}}$

$= \csc^2{x}\tan{x}$

$= \frac{\csc^2{x}}{\cot{x}}$.

3. Hello, nightrider456!

The first one is not an identity.

$\frac{\cos2{x}}{\sin x}\:=\: \frac{\cot^2\!x-1}{\csc x}$

Start with the left side: . $\frac{\cos2x}{\sin x} \;=\;\frac{\cos^2\!x-\sin^2\!x}{\sin x}$

Divide top and bottom by $\sin^2\!x:\;\;\frac{\dfrac{\cos^2x}{\sin^2x} - \dfrac{\sin^2x}{\sin^2x}}{\dfrac{\sin x}{\sin^2x}}$

And we have: . $\frac{\left(\dfrac{\cos x}{\sin x}\right)^2 - 1}{\dfrac{1}{\sin x}} \;=\;\frac{\cot^2\!x - 1}{\csc x}$

4. Sorry, fixed it now thanks for having a look

5. Originally Posted by nightrider456
Have an assignment due in the near future, and these four are giving me a hassle to prove, any help or advice would be appreciated.

Prove
$
\frac{1+tan{x}}{sinx}= cscxsecx
$

$
tanx+cotx=\frac{csc^2{x}}{cotx}
$
$\frac{1 + \tan{x}}{\sin{x}} = \frac{1}{\sin{x}} + \frac{\tan{x}}{\sin{x}}$

$= \csc{x} + \frac{\frac{\sin{x}}{\cos{x}}}{\sin{x}}$

$= \csc{x} + \frac{1}{\cos{x}}$

$= \csc{x} + \sec{x}$...

Are you sure you weren't supposed to prove it's equal to $\csc{x} + \sec{x}$ instead of $\csc{x}\sec{x}$?