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
$\displaystyle
\frac{1+tan{x}}{sinx}= cscxsecx
$
$\displaystyle
tanx+cotx=\frac{csc^2{x}}{cotx}
$
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
$\displaystyle
\frac{1+tan{x}}{sinx}= cscxsecx
$
$\displaystyle
tanx+cotx=\frac{csc^2{x}}{cotx}
$
$\displaystyle \tan{x} + \cot{x} = \frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}$
$\displaystyle = \frac{\sin^2{x}}{\sin{x}\cos{x}} + \frac{\cos^2{x}}{\sin{x}\cos{x}}$
$\displaystyle = \frac{1}{\sin{x}\cos{x}}$
$\displaystyle = \frac{\sin{x}}{\sin^2{x}\cos{x}}$
$\displaystyle = \frac{1}{\sin^2{x}}\cdot \frac{\sin{x}}{\cos{x}}$
$\displaystyle = \csc^2{x}\tan{x}$
$\displaystyle = \frac{\csc^2{x}}{\cot{x}}$.
Hello, nightrider456!
The first one is not an identity.
Please check for typos.
$\displaystyle \frac{\cos2{x}}{\sin x}\:=\: \frac{\cot^2\!x-1}{\csc x}$
Start with the left side: .$\displaystyle \frac{\cos2x}{\sin x} \;=\;\frac{\cos^2\!x-\sin^2\!x}{\sin x}$
Divide top and bottom by $\displaystyle \sin^2\!x:\;\;\frac{\dfrac{\cos^2x}{\sin^2x} - \dfrac{\sin^2x}{\sin^2x}}{\dfrac{\sin x}{\sin^2x}} $
And we have: .$\displaystyle \frac{\left(\dfrac{\cos x}{\sin x}\right)^2 - 1}{\dfrac{1}{\sin x}} \;=\;\frac{\cot^2\!x - 1}{\csc x} $
$\displaystyle \frac{1 + \tan{x}}{\sin{x}} = \frac{1}{\sin{x}} + \frac{\tan{x}}{\sin{x}}$
$\displaystyle = \csc{x} + \frac{\frac{\sin{x}}{\cos{x}}}{\sin{x}}$
$\displaystyle = \csc{x} + \frac{1}{\cos{x}}$
$\displaystyle = \csc{x} + \sec{x}$...
Are you sure you weren't supposed to prove it's equal to $\displaystyle \csc{x} + \sec{x}$ instead of $\displaystyle \csc{x}\sec{x}$?